Systems of Linear Equations
3.1 Substitution Method for a Linear System
3.2 Systems of Linear Inequalities
3.3 Characteristics of a System of Linear Inequalities
3.4 Solving Systems of Linear Equations with More than Two Unknowns:
Elimination Method
The summary will review the most important ideas, not necessarily all the terms that were
bolded in the body of the section.}This chapter covers a very important class of mathematical
problems. Here, you will learn to find the values of two or more variables at the same time.
These problems occur frequently in business and economics, as well as in the social sciences,
life sciences, and physical sciences. A manufacturer, for example, may need to know how
much can be produced of two different products using information about the available
amounts of two different raw materials. A marketing manager may want to find out how
effective her advertising is in two different magazines, based on information about the
responses from two months of advertising. We’ll begin our investigation by looking at a
classic example of diet design, based on consumption of only two foods, for simplicity.
As we shall see in section 3.1, the information that comes with these problems will enable
us to write two linear equations that describe the relationships between these variables. The
two equations will comprise a system of simultaneous equations. For problems with two
variables and two equations, the solution takes only a few steps. Use this section as an
opportunity to practice “seeing” the linearity in the description of a problem and to write the
equations that match the description.
Sections 3.2 and 3.3 explore the topic of linear inequalities. When a linear equation is
written with a less than (<) or greater than (>) sign rather than an equal (=) sign, the solution is
not a line but an entire half-plane in the coordinate system. Linear inequalities have important
applications in business problems where a decision maker is trying to find a combination of
variables that maximizes profit or minimizes cost, subject to certain restrictions on the
variables.
Section 3.4 returns to the topic of simultaneous linear equations but considers problems
that have three or more variables. The simple method of section 3.1 does not work as easily on
these larger problems. Fortunately, an alternative method, the “elimination method,” does
work well and can even be performed by spreadsheets and graphing calculators. But you
should still focus on learning how to read an application problem and to write out the system
of equations that describes it. The application exercises are good practice for this.
3.1 Substitution Method for a Linear System
Learning Objectives
Write a system of linear equations corresponding to a verbal
description.
Solve a system of linear equations graphically.
Solve a system of linear equations algebraically by the substitution
method.
Solve applied problems using tables, graphs and systems of equations.
Chapter 3
Systems of Linear Equations and Inequalities
2
Spotlight on Designing a Diet
An important task in hospitals, the military, and some college food service operations is to
create a menu that fulfills specific dietary requirements. In a simplified version of this
problem, suppose that a dietitian is trying to create a meal that provides a person’s total daily
requirement of vitamin C and iron by serving exactly two dishes—potatoes and spinach.
For a female between the ages of 19 and 24, the recommended daily allowance of vitamin
C is 60 milligrams (mg) and that of iron is 15 mg. The dietician needs to determine how many
potatoes and how much spinach a 21-year-old female must eat per day to fulfill these
requirements.
{INSERT Photo 3.1}
When two linear equations must be solved at the same time, they constitute a system of
simultaneous equations. To solve the system of equations, we must find values of both X and
Y that simultaneously satisfy both equations. There are two methods for solving such
problems, one graphical and the other algebraic. We begin with an example of the graphical
approach.
A GRAPHICAL SOLUTION: — INTERSECTION OF TWO LINES
To solve a system of equations, we plot each equation on the same coordinate grid and
estimate the point of intersection. This employs the key insight needed to solve two equations
simultaneously: in order for a point to satisfy both equations, it must lie on both lines. Only a
point at the intersection of the two lines can satisfy both equations simultaneously.
Example 1
Solving a System of Two Equations Graphically
Solve the following two equations simultaneously by the graphical method:
A. 2X + 5Y = 20,
B. 4X – 3Y = 14
Solution
Linear equations in the general form are easy to graph because with a simple calculation
one can find the vertical and horizontal intercepts. Starting with equation A, 2X + 5Y = 20, we
find the vertical intercept by setting X = 0 and substituting that into the equation:
2(0) + 5Y = 20,
5Y = 20,
Y = 4.
So the vertical intercept is the point with coordinates (0, 4).
Similarly, we find the horizontal intercept for equation A by setting Y = 0 and solving,
2X + 5(0) = 20,
2X = 20,
X = 10.
So the horizontal intercept is (10, 0).
We graph the line 2X + 5Y = 20 by plotting the points (0, 4) and (10, 0) and connecting
the points with a straight line.
By repeating the steps above for equation B, we determine its vertical and horizontal
intercepts. Using the equation 4X – 3Y = 14, we find the intercepts to be (0, <->14/3) and (7/2,
Chapter 3
Systems of Linear Equations and Inequalities
3
0), respectively. Again, we graph the line by plotting these points and connecting them with a
straight line.
The graph of these two equations appears in Figure 3.1.1. The solution to the system of
equations is the point where the two lines intersect. By inspection, it appears to have an Xcoordinate of about 5 and a Y-coordinate of about 2. Thus, the solution to the system of
equations should be close to X = 5, Y = 2.
{INSERT FIG. 3.1}
Learning with Technology
Finding the Intersection of Two Lines
One of the special features of a graphing calculator is its capability of finding the
intersection of two lines or curves. Although the graphing calculator is optional with
this textbook, a word here about its use will help some students.
A graphing calculator can plot several equations at once, but each must be given to
the calculator in the functional form “Y = . . . .” To obtain a graphical solution to the
system of equations above using a graphing calculator, we must rewrite the equations
in this fashion. The first equation in Example 1 was 2X + 5Y = 20. We subtract 2X
from both sides and divide through by 5 to get Y = 4 – 0.4X. Doing the same with the
second equation, taking care when dividing through by <->3, the system becomes
A. Y = 4 – 0.4X,
B. Y = <->14/3 + 4/3X.
Now you are ready to use the graphing calculator, following these steps:
1. Enter these two equations as Y1 and Y2 or on any two other available lines in the Y
= menu. Clear any other equations that might have been selected for plotting the last
time you used the calculator and select these two alone.
{INSERT FIG. UN3.1a}
2. Find a graphing window that will display these lines and their intersection. The
slope-intercept forms in A and B reveal the vertical intercepts of each graph. A
vertical screen interval of [<->5,4] will show both intercepts.
What horizontal screen interval will display a complete graph? Setting Y = 0 in each
equation we do a quick calculation, as in Example 1, to find the horizontal intercepts:
X = 10 for the first equation and X = <->7/2 for the second. The horizontal viewing
window should therefore extend as far as 10 to the right. The vertical intercepts
correspond to X = 0, so all intercepts will appear if we use a horizontal window
[0,10].
{INSERT FIG. UN3.1b}
3. Make sure that both equations are selected for graphing, and then call up the graph.
It is the same as Figure 3.1, but it should appear on the calculator as shown here.
Most calculators do not show the numbers on the axes, but you can infer them from
the tick marks and the window boundaries.
{INSERT FIG. UN3.1c}
4. Most graphing calculators can find the intersection of two lines. A CALC
INTERSECT routine in the Texas Instruments models will find the intersection.
{INSERT FIG. UN3. 1d}
Chapter 3
Systems of Linear Equations and Inequalities
4
If your calculator does not have an INTERSECT routine, the TRACE feature will
still allow you to move along one line until you get to the intersection with the other
line. Toggle back and forth between the two lines until you find the point where the
Y-values for each equation are as close as possible. ZOOM in on the point of
intersection and TRACE to get on one of the lines and find a more accurate estimate.
These show that the solution to the system of equations is X = 5 and Y=2.
{INSERT FIG. UN3.1e}
AN ALGEBRAIC SOLUTION: THE SUBSTITUTION METHOD
An algebraic way to solve a system of linear equations is called the substitution method. This
will give a more precise solution than the graphical method above. We’ll use this method to
solve the problem stated in Example 1 above.
Example 2
Solving a System of Two Equations Algebraically
Solve the system from Example 1 algebraically:
A. 2X + 5Y = 20,
B. 4X – 3Y = 14.
Solution
1. Pick one of the two equations and solve for one of the variables. The two equations in
Example 1 are both written in the general form, so it doesn’t matter which one we pick or
which variable we solve for. Here we’ll begin by solving for X in equation A.
Starting with
2X + 5Y = 20,
we subtract 5Y from both sides,
2X = 20 – 5Y,
and divide through by 2,
X = 10 –
5
Y.
2
2. Use this formula for X in the other equation. We substitute the formula 10 –
place of X in equation B and then solve for Y:
4X – 3Y = 14,
4(10 –
5
Y) – 3Y = 14,
2
40 – 10Y – 3Y = 14,
<->13Y = <->26,
Y = 2.
This gives us one part of the solution, Y = 2.
5
Y in
2
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Systems of Linear Equations and Inequalities
5
3. Substitute this known value for Y into either equation A or B, and calculate the
corresponding value of X. Equation A was already put into slope intercept form for X. Using
that we get
X = 10 –
5
Y,
2
X = 10 –
5
(2),
2
X = 5.
The answer X = 5 and Y = 2 was suggested as an approximate solution in our graphical
analysis of Example 1, which shows that the graphical solution can be a good double check of
our calculations in the algebraic method. To verify the solution algebraically, we substitute the
values for X and Y into equation B:
4X – 3Y = 14,
4(5) – 3(2) = 14 .
Make a Note
Substitution Method in Three Steps
1. Pick one of the two equations and solve for one of the variables.
2. Substitute the formula for that variable into the other equation wherever
that variable appears and solve for the remaining variable.
3. Substitute the value you obtained in Step 2 into the equation you got in
Step 1 and calculate the value of the first variable.
Example 3
Designing a Diet
Let’s return to the application introduced in the Spotlight at the beginning of this chapter. The
problem was to design a diet consisting of only two foods, spinach and potatoes. The objective
of the diet is to satisfy two minimum daily requirements: 60 mg of vitamin C and 15 mg of
iron.
To plan the diet, we need to know how much of each nutrient is found in a pound of each
type of food. The table shows the nutritional values of each. *
Nutrient
Vitamin C (mg)
Iron (mg)
Per
pound of
. spinach
45
16
Per pound
potatoes
58
1.5
The quantities of spinach (S, lbs.) and of potatoes (P, lbs.) that fulfill the vitamin C
requirement should satisfy an equation that reads as follows:
The amount of vitamin C from spinach and the amount from potatoes must sum to 60 mg.
Each pound of spinach gives 45 mg of vitamin C, and each pound of potatoes gives 58 mg, so
the statement above translates into mathematical form as
*Nutritive Value of Foods, Home and Garden Bulletin, No. 72 (Washington, DC: US Government
Printing Office). < http://www.nal.usda.gov/fnic/foodcomp/Data/HG72/hg72.html>
Chapter 3
Systems of Linear Equations and Inequalities
6
A. 45S + 58P = 60.
By the same logic, we may also write an equation that expresses the iron requirement of
15 mg using the data on iron content from the table,
B. 16S + 1.5P = 15.
How can we find a value for S and a value for P that satisfy both of these equations
at the same time?
Solution
Graphical Solution
To get started on such problems, it always helps to draw a picture. Figure 3.2 shows a graph of
both equations on the same chart. We do so by plotting the vertical and horizontal intercepts
of each equation and connecting the points by a straight line as we did in Example 1. The
intercepts (S, P) of equation A are (0, 60/58) and (60/45, 0). Those of equation B are (0, 10)
and (15/16, 0).
{INSERT FIG 3.2}
Think about This
Examine Figure 3.2, and see if you can answer this question yourself: Are
there any combinations of S and P that can satisfy both equations at the
same time?
Remember that only at the point of intersection are both equations satisfied simultaneously. In
Figure 3.2, it looks as though the coordinates of the point of intersection are approximately S
= 0.9 pounds of spinach and P = 0.5 pounds of potatoes. But an eyeball guess about the
coordinates of the intersection point won’t be precise enough for a dietician, who is
responsible for meeting dietary requirements. So let’s see how to get an exact answer.
Algebraic Solution
The first step is to isolate one of the variables. Neither the
vitamin C nor the iron equation is in slope-intercept form,
so it does not matter which variable we isolate or in which
equation. We shall isolate the variable S in the vitamin C
requirement, equation A,
A. 45S + 58P = 60.
Subtract 58P from both sides,
45S = 60 – 58P.
Divide through by 45,
S=
60
58
–
P,
45
45
S = 1.333 – 1.289P.
Make a Note
Using Decimals and Rounding Answers
In the real world, the solutions to math problems
rarely come out to be integers or simple
fractions. They are real numbers, which we
approximate as decimals. It is the common
practice of applied mathematicians to round their
answers to the same number of significant digits
as the data given in a problem, except where an
answer calls for dollars and cents. We will follow
that practice here. However, during the
calculations toward a solution, we will carry at
least one extra digit. Otherwise, several small
rounding errors might add up to a larger error in
the final solution.
The next step is to substitute this formula wherever S
appears in the iron requirement, equation B, and then solve for the remaining variable, P. We
start with the iron equation as it was originally stated,
B. 16S + 1.5P = 15.
Now, substitute the formula 1.333 – 1.289P for the variable S in this equation,
16(1.333 – 1.289P) + 1.5P = 15.
Chapter 3
Systems of Linear Equations and Inequalities
7
Think about This
Be sure you understand how this equation was created. It is a combination
of equations A and B above, formed by putting the formula for S from
equation A into the place where S was in equation B.
Expand the parentheses using the distributive rule,
21.333 – 20.624P + 1.5P = 15,
and combine the terms in P,
<->19.124P = <->6.333.
Divide through by <->19.124, and retain two significant digits in the answer (because the data
in the problem were stated in two significant digits),
P ≈ 0.33 pounds of potatoes.
This is the value of one variable. Now we must find a value for the other variable.
Think about This
Knowing P now, how can we find the corresponding value of S?
For this purpose, we go back to the first equation. We substitute this value for P into the
version of equation A in which the variable S was isolated:
S = 1.333 – 1.289P,
S = 1.333 – 1.289(0.33),
S = 1.333 – 0.425,
S ≈ 0.91 pounds of spinach (to two decimal places).
The solution to the problem is that a daily diet of 0.91 pounds of spinach and 0.33 pounds
of potatoes will provide the required amounts of both vitamin C and iron.
There are two ways to check the correctness of this solution, one graphical and one
algebraic. The graphical way to check the answer is to sketch a plot of the two equations and
see if the intersection of the lines appears to be near P = 0.33 and S = 0.91. For this we may
refer back to our graphical solution in Figure 3.2. The lines do appear to intersect near S = 0.9
and P = 0.3.
The algebraic way to check this solution is to substitute these values of S and P into each
of the two original equations. We want to see if the calculation results in the appropriate righthand-side value of the equation—at least to within a tolerable error (third significant digit) due
to rounding off our divisions in the process of calculation. Start with the vitamin C equation:
A. 45(0.91) + 58(0.33) = 60?
40.95 + 19.14 = 60?
59.99 ≈ 60; close.
Now check the iron requirement:
(b) 16(0.91) + 1.5(0.33) = 15?
14.56 + 0.495 = 15?
15.055 ≈ 15; close.
We conclude that to get the daily required amounts of vitamin C and iron on a diet of
potatoes and spinach, a woman aged 19-24 would need to eat about 0.91 pounds of spinach
and 0.33 pounds of potatoes per day.
There are numerous important applications of systems of linear equations in business and
economics. Let’s take a look at some of these.
Chapter 3
Systems of Linear Equations and Inequalities
8
Example 4
Making Production Decisions in a Wood Shop
Let’s say that a wood shop produces only two products, bookcases and chairs. These two
outputs require the same two inputs, labor time and machine time. This table gives the
necessary information about how much of each input is used to create one unit of each
product, for example, that the manufacture of one bookcase requires 0.75 labor hours and 0.5
machine hours.
Input requirements
Per bookcase
Per chair
Labor
0.75
2.5
Machine
0.5
1.0
The owner of the workshop has 2 employees who each work a 40-hour week. In addition,
his machinist runs the machine for 36 hours a week and does maintenance on the machine for
the remainder of the 40-hour workweek. How many bookcases and chairs can the shop make
if it uses all the labor and machine time available?
Solution
FORMULATING THE PROBLEM
To set up this problem mathematically, we write two equations in two unknowns. (See Fig.
3.3). The first equation will say in words, “The number of hours needed to make B bookcases
plus the hours needed for C chairs must equal 80 hours.” The second is analogous for machine
hours.
A. 0.75B + 2.5C = 80 (labor hours)
B. 0.5B + C = 36 (machine hours)
{INSERT FIG. 3.3}.
ALGEBRAIC SOLUTION
1. Isolate one of the variables in one of the equations. In the machine equation B, the variable
C already has a coefficient of 1, so that makes our task a little easier:
B. 0.5B + C = 36 (machine hours).
Subtract 0.5B from each side:
C = 36 – 0.5B.
2. Substitute this expression for C into the labor equation A and solve for the remaining
variable:
A. 0.75B + 2.5C = 80 (labor hours),
0.75B + 2.5(36 – 0.5B) = 80,
0.75B – 1.25B + 90 = 80,
<->0.5B = <->10,
B = 20 bookcases.
3. Substitute this value for B into the isolated version of equation B:
C = 36 – 0.5(20),
C = 36 – 10,
C = 26 chairs.
Chapter 3
Systems of Linear Equations and Inequalities
9
In conclusion, when using all the labor hours and machine time available, namely 80
labor hours and 36 machine hours, the shop can make 20 bookcases and 26 chairs. We check
these solutions by substituting them into the original equations:
A. 0.75B + 2.5C = 80,
0.75(20) + 2.5(26) = 80?
15 + 65 = 80?
80 = 80 exactly.
B. 0.5B + C = 36,
0.5(20) + (26) = 36?
10 + 26 = 36?
36 = 36 exactly.
Learning with Technology
Find the size of the window needed to view the graphs of equations A and B.
Put equations A and B in slope intercept form as “C = . . . ,” but keep the constants in
fractional form. Hint: Equation A would be written as C = 80/2.5 – 0.75/2.5*B.
Graph the equations and find their intersection using the INTERSECT or TRACE
routine.
How would the solution change if there were 40 machine hours available rather than
36? What if there were 50? What if 60?
Application
Market Supply and Demand
The “market” for any product consists of buyers and sellers. The buyers’ need for the product
is described by a demand curve (or demand equation). This is a graph that shows the
relationship between the price (P) in the market and the quantity (Q) of the good that would be
demanded (purchased) by buyers. It is a general economic principle that when the price is
higher, people consume less of a product. Graphically, this means that the demand curve
slopes downward.
The supply side of a market is described by a supply curve (or supply equation) that
indicates what quantity (Q) would be produced by suppliers and offered in the market at each
level of price (P). The supply curve slopes upward because producers require a higher price if
they are to increase production output within the capacity of their existing factories. The
reason is that when a factory has to produce a lot more units of a product in the same amount
of time, crowding, mistakes, and inefficiencies tend to result in a higher cost of production.
At a price where the quantity demanded is equal to the quantity supplied, the market is
said to have “cleared.” In other words, the combination of price and quantity in the market has
to fit both the buyers’ demand equation and the producers’ supply equation. The combination
of price and quantity that makes this possible is called the market equilibrium. In other
words, the P and Q in the market equilibrium problem have to satisfy two different equations
at the same time.
Example 5
Demand for CD Players
Many electronics companies sell CD players on the Internet. Suppose that in their experience,
customers will not pay more than $200 for a particular model of CD player. But customers
would be willing to buy a player (in any 1 year) if the price is lowered. Let’s say that to sell
Chapter 3
Systems of Linear Equations and Inequalities
10
one more player per year, no matter how many players are currently selling, the price must
come down by $0.006.
Create the demand equation that describes the relationship between P and Q and plot it.
Solution
According to this description, there will be a linear relationship between quantity sold in the
entire market and the price of a CD player. The demand line will have a vertical intercept of
$200 (where Q = 0) and a slope of <->0.006. The equation representing this demand curve
would appear in slope-intercept form as
P = 200 – 0.006Q.
Using this market demand equation, you can calculate the corresponding value of P if you
know the number Q of players that need to be sold in a year. Likewise, given a known value
for P, you can solve for the corresponding value of Q to find out how many players will be
sold at that price.
Practice
1. If 10,000 players need to be sold in a year, what price should the retailers
set?
2. How many players would customers want to buy if the price is $100?
Answers: (1) Use the demand equation P = 200 – 0.006Q with Q = 10,000, getting P = $140.
(2) Solve the demand equation P = 200 – 0.006Q with P = 100, and find Q ≈ 16,667 players.
The market demand equation describes a negative association between P and Q. The
higher the Q, the lower the P. This is seen in the downward-sloping graph of Figure 3.4, and it
is evident from the negative slope coefficient in the equation P = 200 – 0.006Q.
{INSERT FIG 3.4}
The supply equation tells a very different story about P and Q. In many industries, the
production cost per unit tends to increase as the production level increases, due to production
inefficiencies, overtime pay to workers, and so on. Thus, manufacturers will supply more of a
product only if the price is higher. In other words, the higher the Q, the higher must be the P.
This is characteristic of a positive association between the variables P and Q. We should
expect that the supply equation will have a positive slope coefficient and that its graph will
have an upward slope.
Another feature of the production process is that, from the manufacturer’s point of view,
there is a minimum price below which producing new players is not profitable, so no players
will be produced. In graphical terms, this means that the supply line will have a vertical
intercept at some positive value of P.
Example 6
Supply of CD Players
Suppose that in our CD player example it costs the manufacturers of these players a minimum
of $50 to produce each player. For each additional batch of 1,000 players that they have to
produce in a given year, the production cost per player goes up by $4, meaning $0.004 per
player. The production cost is the minimum price at which the manufacturer would be willing
to supply the players.
Write an equation that describes the manufacturer’s willingness to supply CD players and
plot it.
Chapter 3
Systems of Linear Equations and Inequalities
11
Solution
Again, we have a linear relationship between P and Q. The supply line will have a vertical
intercept at $50, where Q = 0, and a slope of 0.004. Thus, the equation describing the
manufacturers’ price in relation to the number of players produced can be written as
P = 50 + 0.004Q.
Just as with the demand equation, if you are given a value for Q, you can calculate the
corresponding P very quickly. Or, given a value for P, you can solve for the corresponding
value of Q rather easily. But the supply equation describes a very different relationship
between P and Q. As we see in Figure 3.5, it has an upward slope. Recall that the demand
equation in Figure 3.4 sloped downward.
{INSERT 3.5}
In the Practice box, we calculated that customers would purchase 16,666 units at a price
of $100. Now let’s find out how many players the manufacturers would be willing to produce
if the price is $100. We solve the supply equation, P = 50 + 0.004Q with the value of P set at
$100,
100 = 50 + 0.004Q,
50 = 0.004Q,
Q = 12,500 players.
These two calculations show that at a price of $100 per player, there is a demand for 16,666
units, but the manufacturers are willing to produce only 12,500 units. When a market has an
excess of demand over supply (or when supply exceeds demand) it is said to be in
disequilibrium.
What combination of price and quantity will balance supply and demand? It will be a
level of P and a level of Q that satisfies both the demand equation and the supply equation
simultaneously. Graphically, the market equilibrium will be a point with coordinates (P, Q)
that lies simultaneously on the market demand line and on the supply line. That is the point of
intersection of the two lines.
Example 7
Market Equilibrium of CD Players
Determine the point at which demand meets supply in our CD player application.
Solution
GRAPHICAL SOLUTION
By plotting both the demand and supply equations from the previous examples on the same
graph, we can see that the equilibrium will correspond to Q ≈ 15,000 players at a price of P ≈
$120 each in Figure 3.6.
{INSERT FIG. 3.6}
ALGEBRAIC SOLUTION
Algebraically, the market equilibrium point (P, Q) is the pair of numbers that satisfy
simultaneously the demand and supply equations,
A. P = 200 – 0.006Q (demand equation),
B. P = 50 + 0.004Q (supply equation).
Chapter 3
Systems of Linear Equations and Inequalities
12
To find the equilibrium algebraically, we solve this system of two equations
simultaneously, and for that we use the substitution method. Each of these equations is already
in slope-intercept form, so they have one variable isolated on the left side of the equality sign.
It does not matter which equation we use for the substitution. Let’s take the demand equation.
We substitute the formula 200 – 0.006Q everywhere the letter P occurs in the supply
equation,
P = 50 + 0.004Q (supply equation).
This gives us a new equation,
(200 – 0.006Q) = 50 + 0.004Q.
We now solve this for the unknown value of Q. Add 0.006Q to both sides:
200 = 50 + 0.010Q.
Subtract 50 from both sides:
150 = 0.010Q.
Divide through by 0.010 to get Q:
Q = 15,000.
To find the corresponding value of P, we may use either equation A or B. Substitute the
value Q = 15,000 into it, and solve for P. We take equation A:
P = 200 – 0.006(15,000),
P = 200 – 90,
P = $110.
In summary, the solution to the system of equations is the pair of values Q = 15,000 and P
= $110 as Figure 3.6 confirms. In other words, the point where the two lines intersect in the QP plane has coordinates (15,000, 110). This point corresponds to the production and
consumption of 15,000 CD players at a market price of $110 each.
PARAMETER ESTIMATION
To write a linear equation, we have been using large letters for the variables and small letters
for the constants or parameters, which will be specific numbers when the equation is actually
used. The slope-intercept form of the linear equation,
Y = mX + b,
has two parameters: the coefficient of X (the slope parameter, m) and the constant term, b.
In most of the application problems we encounter, the linear equation describes a general
rule that shows the relationship between the variables. Giving known values to the parameters
in the equation makes that relationship specific, as when we define a particular demand
equation to be
Y = 200 – 0.006X.
As you have seen, the task in most problems is to find a value for an unknown variable when
we already know the parameters of the equation.
However, one of the most important applications of systems of equations is to find the
parameters that define a specific relationship between two variables. To find the two
parameters for a linear equation, we have to use information about the relationship between
the two variables in two or more observed cases. In the applications we will examine below,
we will use exactly two cases. Two cases give us two points of data, and two points determine
a line. This is enough to give us a specific estimate of the two parameters in a linear equation.
Chapter 3
Application
Systems of Linear Equations and Inequalities
13
A Formula for the Price of a House
Two of the most important factors in determining the price of a house are its square footage of
floor space and the amount of land that comes with it. In other words, the price P of a house is
a function of its square footage S and the amount of its land, L.
In general, the price of a house will go up linearly with its square footage and likewise
with the area of land on which the house sits. This means that the general form of an equation
for the price will be
house price P = cS + dL,
for some parameters c and d. The constants, or parameters as they are often called, indicate
respectively the number c of dollars paid per square foot and the number d of dollars paid per
acre.
When people buy a house and land, it is typically bought as one package and paid for
with one sum of money. But the tax laws treat the building and the land very differently. If
you buy a house in order to rent it out, you can depreciate part of the value of the building
every year (that is, treat it as a business expense), but you cannot depreciate the value of the
land. The logic of the Internal Revenue Service is that buildings wear down over time, so it is
reasonable to allow businesspeople a tax deduction for that, but land does not wear down.
Therefore, anyone who buys a house to rent it out wonders, “How much of the purchase price
is attributable to the house and how much to the land?” This question can be answered by
setting up a system of two equations, each of which has two unknowns.
This problem has an interesting twist, like the problem of finding a commission rate for a
salesperson. The unknowns in the system of equations are going to be the constants c and d in
the house price formula, and the constants are going to be the actual characteristics of two
houses that have sold for known prices. If you can mentally tolerate this turn of the tables for
another minute, you’ll see how useful this idea is.
Example 8
Price of Land and Buildings
Suppose that we have real data on two houses that recently sold in town. The data are
summarized here in the table.
Property
Size of the building
Land
Market price
(sq. ft.)
(acres)
110 Main St.
900
0.25
$105,000
412
St.
1,200
0.40
$145,000
Dexter
As surprising as it may seem, this is all the information we need in order to determine
how the price of a property is calculated from the square footage of the house and the acreage
of the land. Using this table, write two equations in the two unknown parameters.
Solution
It is often helpful to first look for an equation that can be stated in words. In this example, we
will set up two equations that each state,
The cost of the building plus the cost of the land equals the cost of the property.
Assuming that in the town, the general price of the building is c dollars per square foot, then
the cost of the building on property A must be c  900. Likewise, if the cost of land is d dollars
per acre, then the 0.25 acres at 110 Main Street must have cost d  0.25. The sum of these must
equal the price paid for that property. This logic gives us the first of the following equations.
The second equation is obtained similarly:
Chapter 3
Systems of Linear Equations and Inequalities
14
A. c900 + d  0.25 = 105,000 (price of 110 Main St.),
B. c1,200 + d  0.40 = 145,000 (price of 412 Dexter St.).
Notice what we have done here. In each equation, we used actual data for the “variables”
S and L but left the parameters c and d unspecified. This gives us a linear equation in which
the unknowns are the parameters c and d. This is an example of how real data are used to
estimate the parameters of a mathematical model (an equation) for a phenomenon like house
prices.
We can solve these two linear equations using the substitution method. Let’s take the first
equation and isolate the unknown d.
Starting with
c  900 + d  0.25 = 105,000,
we get
d = 420,000 – c  3600.
Substitute this into the second equation, and solve for the unknown parameter c:
c1200 + (420,000 – c  3600)0.40 = 145,000.
Expand the parentheses,
1,200c – 1,440c + 168,000 = 145,000.
Combine like terms,
<->240 c = <->23,000,
and divide through,
c ≈ $95.83 per square foot.
Our equation for the parameter d was
d = 420,000 – c =3600.
Substitute the known value for c back into our equation for the parameter d and solve.
d ≈ 420,000 – (95.83)3,600.
d ≈ $75,012 per acre.
There we have it. The actual data for two properties have enabled us to determine the two
unknown parameters of a linear equation that describes the price of a property as a function of
size of the building and the amount of the land.
To check these answers graphically, we plot the two original equations and see if the
intersection corresponds with the solutions we just found. Figure 3.7 shows that we should
expect a solution that has a price per acre somewhat below $100,000 and a price per square
foot a little below $100. The solution c = $95.83 per square foot and d = $75,012 per acre is
consistent with this picture.
{INSERT FIG. 3.7}
To check the answers algebraically, we substitute our values of c and d into the original
equations and see if the equations hold up. The equation for 110 Main Street is
A. c  900 + d  0.25 = 105,000 (price of 110 Main St.).
Substitute the solutions c = 95.83 and d = 75,012 into this equation and see if the left and right
sides match up:
(95.83)  900 + (75,012)  0.25 = 105,000,
86,247 + 18,753 = 105,000,
Chapter 3
Systems of Linear Equations and Inequalities
15
105,000 = 105,000. [ED: code?]
Now do the same for 412 Dexter Street,
B. c  1,200 + d0.40 = 145,000,
(95.83)  1200 + (75,012)0.40 = 145,000,
114,996 + 30,004.80 = 145,000.
145,000.80 ≈ 145,000 (close enough).
Our purpose in searching for Please restore all in this sentence}the costs per square foot
and per acre was to write an equation for the cost of a property if the size of the building and
the amount of the land were unknown. We have found those cost parameters, so we write
them into the template for our linear equation for the property price,
P = $95.83S + $75,012L,
where S is the size of the house, and L is the amount of land.
Using this equation, we can now determine the separate value of the building and of the
land in each of the houses for which we have data. Consider the property at 110 Main St. For
tax purposes, the value of the building would be $95.83(900) = $86,247, and the value of its
land would be $75,012(0.25) = $18,753.
It is quite remarkable that a few points of data enable us to tease out information that
seemed at first to be buried inextricably in the situation. You will find that to be true in the
examples that follow.
Application Estimating Productivity Coefficients
A production manager in a firm needs to know how long it takes employees to produce
different products. One way would be to observe employees in the process of manufacturing
each product and to measure the time they take on each type, but employees do not usually
like to be watched.
There is another way, though, which does not require someone to observe people working
on each type of product individually. A manager can simply count the number of units of each
type of product produced in, say, an 8-hour day. If there are only two products and the
manager has this information on two days, it will be possible to calculate the time it takes to
produce one unit of each type of product.
How can a manager do this? Let’s assume that there are two products, labeled 1 and 2
and that there is an unknown amount of time required to produce each product. We write c1 for
the number of hours needed to manufacture a unit of product 1 and c2 for the number of hours
needed to manufacture a unit of product 2.
The total time that will be spent in manufacturing a quantity Q1 of product 1 will be the
number of hours required to manufacture a single unit multiplied by the number of units
produced. That total is c1Q1. Likewise, the total time that will be spent in manufacturing a
quantity Q2 of product 2 will be c2Q2.
The mathematical model for the total number of hours (H) a worker needs to manufacture
Q1 units of product 1 and Q2 units of product 2 is therefore expressed by the equation,
H = c1Q1 + c2Q2, otherwise the variable H is defined but note used? OK}
Example 9
Employee Productivity
Let’s introduce some values to make the application concrete. Suppose we know the
following:
Chapter 3
Systems of Linear Equations and Inequalities
16
Day 1: During this 8-hour day an employee made 6 units of product 1 and 8 units of
product 2.
Day 2: During this 8-hour day, the same employee made 3 units of product 1 and 12 units
of product 2.
We will write one equation using each day’s information. In plain English, the equations will
state:
The number of hours used to manufacture product 1 plus the number of hours used to
manufacture product 2 is equal to the 8 hours in a worker’s day.
In mathematical form, this reads as
A. c1  6 + c2  8 = 8 hours worked in the first day,
B. c1  3 + c2  12 = 8 hours worked in the second day.
Find out how much time it takes to make 1 unit of each product by solving this system of
simultaneous linear equations.
Solution
GRAPHICAL SOLUTION
These two equations are in general form, so it should be easy to do so in paper and pencil by
finding the vertical and horizontal intercepts of each line and then drawing the lines.
The intercepts of line A are at (0, 1) and (4/3, 0). The intercepts of line B are at (0, 2/3)
and (8/3, 0). A plot of these two lines on a window [0,3] by [0,1] appears in Figure 3.8.
{INSERT FIG 3.8}
The dashed lines show that the solution to the system of equations ought to be production
times of approximately c1 = 2/3 of an hour per unit of product 1 and c2 = 1/2 hour per unit of
product 2.
ALGEBRAIC SOLUTION
Since we are accustomed to solving systems in which the unknown appears to the right of the
specified coefficient, we may use the commutative law for real numbers to rearrange the order
of multiplication. This makes the system of equations appear as follows:
A. 6c1 + 8c2 = 8 ,
B. 3c1 + 12c2 = 8.
The first step is to isolate one of the unknowns in one of the equations. We choose to
isolate c2 in equation A because when we divide through by the coefficient of c2, which is 8,
we will get familiar fractions and numbers:
A. 6c1 + 8c2 = 8,
8c2 = 8 – 6c1,
c2 =
8 6
– c1,
8 8
c2 = 1 –
3
c1.
4
The second step is to substitute that expression for c2 into equation B.
3c1 + 12c2 = 8,
3c1 + 12(1 –
3
c1) = 8.
4
Chapter 3
Systems of Linear Equations and Inequalities
17
With a little arithmetic and the rules of algebra, we solve this for the unknown c1.
3c1 + 12 – 9 c1 = 8,
<->6c1 = <->4,
c1 =
2
.
3
In the final step, we substitute this value for c1 back into our equation for c2 to find the
remaining unknown,
c2 = 1 –
=1–
=
3  2
 
4  3
2
4
1
.
2
Our solution to these equations reveals that it takes 2/3 of an hour (or 40 minutes) to
produce each unit of product 1 and 1/2 of an hour (or 30 minutes) to produce each unit of
product 2. These are exact solutions, and they correspond well to the eyeball estimates taken
from the graphs in Figure 3.8.
Application Estimating the Effects of Advertising
Marketing managers often complain, “Half my advertising budget is wasted. I know that. The
problem is, I don’t know which half.”
A marketing manager needs to find out exactly what are the effects of an advertising
campaign. However, it is usually not practical to take out one advertisement at a time in order
to see its effects clearly. Typically a manager tries many advertising tactics at once. But then it
can be a real problem to sort out the individual effects of all the separate advertisements and
other marketing tactics.
For that purpose, it helps enormously to know how to solve systems of equations. With
data about the sales of a product over 2 months, as a result of advertisements of 2 different
types, you can usually determine how much of an effect each of the advertisements was
having.
Example 10
Terry’s Water Purifiers
While still in college, Terry Reilly was given the Midwest dealership for a brand of water
purifiers, which he sells from his dorm room as a way to make extra money. His only
advertising has been a series of 30-second radio ads. He runs them on the campus radio
station, WABC, and on a station in town, WXYZ. The number of ads he ran per week and the
results in each of two weeks are summarized in the table.
Week
Ads on
Ads on Units
WABC
WXY
Sold
Z
1
15
20
55
2
10
15
40
Determine the effectiveness of running ads at each station.
Solution
Chapter 3
Systems of Linear Equations and Inequalities
18
With two radio stations and two weeks of data, it should be possible to tease out the separate
effects of the ads. We will assume that for each station there is a proportional relationship
between number of ads run and the number of units sold per week.
To write this idea in an algebraic form, let the letter A represent the number of ads run per
week in WABC and let X be the number of ads run per week at WXYZ. Let c be the number
of units sold per advertisement on WABC and write d for the number of units sold per
advertisement on WXYZ.
The verbal model, in short, says,
The sales from ads in WABC and the sales from WXYZ make up the total number of purifiers
sold.
This can be written algebraically as
cA + dX = total purifiers sold.
We use each of the two weeks’ data for A, X, and the number of purifiers sold in this algebraic
model to give us two equations:
c15 + d20 = 55 (units sold, week 10
c10 + d15 = 40 (units sold, week 2).
This is a system of two simultaneous linear equations in two unknowns, c and d.
GRAPHICAL SOLUTION
To graph these two equations on the same coordinate system, we should estimate the
intercepts of each equation and choose our window appropriately. In rough numbers, for the
first equation, when c = 0, d ≈ 3, and when d = 0, c ≈ 3. Likewise, for the second equation,
when c = 0, d ≈ 3, and when d = 0, c = 4. So an interval of [0,4] should work for c and [0,3]
should work for d.
Figure 3.9 shows the graphs of these equations on that window. The lines corresponding
to the two equations have nearly the same slopes and intercepts, so it is a bit difficult to see
exactly where they intersect. It appears to be near c = 1 and d = 2.
{INSERT FIG. 3.9}
ALGEBRAIC SOLUTION
How should we start solving this system? For step 1, let’s isolate the c in the second equation
because its coefficient, 10, might help keep the numbers simple for a while when we have to
divide through. Subtracting d  15 from both sides, we get
c  10 = 40 – 15  d.
Dividing through by 10 yields an equivalent form of the second equation,
c = 4 – 1.5  d.
As step 2, we substitute this expression for c into the other equation, and solve for d:
(4 – 1.5d)15 + d  20 = 55,
60 – 22.5d + 20d = 55,
5 = 2.5d,
d = 2 sales per ad at WXYZ.
As step 3, we substitute that value of d back into the equation for c,
c = 4 – 1.5  2,
Chapter 3
Systems of Linear Equations and Inequalities
19
c = 1 sale per ad at WABC.
As a check, we put these values for c and d back into the original equations to see if they
produce equality. They are not used for double checks above. If you feel they are necessary,
OK to use "Question:" instead?
(1)  15 + (2)  20 = 55 (units sold, week 1),
15 + 40 = 55;✓
(1)  10 + (2)  15 = 40 (units sold, week 2),
10 + 30 = 40. ✓
This analysis shows that each ad at WABC results in the sale of 1 water purifier, and each ad
at WXYZ results in 2 sales.
PRACTICAL IMPLICATIONS OF THE MODEL
Does this solution mean that Terry should advertise only in WXYZ, where he gets 2 sales per
advertisement? Not necessarily. Terry should want to advertise where he gets the greatest
return on his money, which means the highest sales per dollar of advertising. Therefore, he
has to consider both the effectiveness of the ad (units sold per ad) and the cost of each ad.
Suppose the spots on WABC cost $10 each and the spots on WXYZ cost $25. Which
radio station had the higher ratio of sales to advertising cost?
Terry knows from the previous analysis that each ad at WABC produces 1 sale. This
means that he sells 1/$10 = 0.10 purifiers per dollar of advertising spent there. At WXYZ,
each ad produces 2 purifier sales, so he sells 2/$25 = 0.08 purifiers per dollar spent there. In
this example, WABC produces the higher number of sales per dollar, even though the number
of responses per ad is lower.
Make a Note
Solving a Linear System for Variables or for Parameters
The template for the general form of a linear equation is
aX + bY = c.
This can be made into a specific linear equation in either of two ways.
1. When specific parameters are known, the equation defines a relationship
between the variables X and Y (e.g., the wood shop problem). A system of
two such equations can be solved to find values for X and Y.
2. When the parameters are unknown, specific values for X and Y and for the
right-hand side number c turn the template into a linear equation with
unknown parameters. Data for two combinations of X, Y, and c will lead to
two linear equations that can be solved to find the parameters a and b.
TWO UNUSUAL CASES
In the examples that we have seen so far in this section, the two equations corresponded to
two distinct lines that intersected in a single point. The coordinates of that point of intersection
gave the values of the two variables that solved the system of equations as a whole. This
principle is quite general. The solution to the system of equations is always the set of points in
the plane that lie on both lines.
But does every pair of lines intersect in a single point?
Think about This
Is it possible to draw one or more examples of pairs of lines that do not
intersect in a single point?
Chapter 3
Systems of Linear Equations and Inequalities
20
There are two ways in which the ordinary situation of a single-point solution can fail. We call
these cases “unusual” because they differ from what one would expect in a system of two
linear equations in two unknowns.
An Inconsistent System
A system of equations is called inconsistent if it has no solution. Two linear equations will be
inconsistent when they do not have a point of intersection. In this case, they must correspond
to two parallel lines (Fig. 3.10).
{INSERT FIG. 3.10}
Algebraically, how can we recognize that a system of two linear equations corresponds to
two parallel lines? The key idea is that parallel lines have the same slope. So, when two linear
equations are written in the slope-intercept form, it will be obvious whether or not their slopes
are the same.
To be parallel, the lines must also have different vertical intercepts. The slope-intercept
form will also reveal whether or not the intercept parameters are different. The graph shown in
Figure 3.10 corresponds to a system of equations that, when written in slope-intercept form,
appears as,
Y = 5X + 7,
Y = 5X + 3.
The slope parameter is the coefficient of the X variable, which is 5 in both equations.
Therefore, these equations represent lines that have the same slope. The vertical intercept is
+7 in the top equation and +3 in the bottom equation, showing that the lines are in fact distinct
from each other.
Not all systems of linear equations come to us in slope-intercept form, so it is not always
possible to recognize an inconsistent system of equations by inspection. However, if you try to
work the substitution method on an inconsistent system, you get an interesting result. What
happens? Let’s see by working an example.
Example 11
An Inconsistent System
Consider the following system of two linear equations in two unknowns,
A. 2Y = X – 4
B. Y + 7 = 0.5X.
Try to solve this system using the steps in the substitution method.
Solution
1. Isolate one variable in one of the equations. The bottom equation is closest to slopeintercept form, so let’s isolate the Y by subtracting 7 from both sides. This gives us the
equivalent system of equations.
A. 2Y = X – 4,
B. Y = 0.5X – 7.
2. Substitute the expression for that variable into the other equation. We put (0.5X – 7) in
place of Y in the top equation, and simplify to solve for X. The top equation becomes
A. 2(0.5X – 7) = X – 4.
Chapter 3
Systems of Linear Equations and Inequalities
21
Expand the parentheses,
2  0.5X – 2  7 = X – 4,
and do the arithmetic,
X – 14 = X – 4.
Add 4 to both sides,
X – 10 = X.
Simplify by subtracting X,
<->10 = 0.
The equation is false!
Notice what happened here. The substitution method led us to an equation that is always
false. That is the algebraic signal that the original system of equations was inconsistent. There
is no combination of X and Y that can solve both equations at the same time.
A Dependent System
One more unusual case is possible for a system of linear equations. We ordinarily do not
expect more than one point of intersection because two lines should intersect in a single point.
But sometimes a system of two linear equations in two unknowns can intersect in an entire
line.
How could this happen? We saw in Chapter 2 that there are many ways to write an
equation for a particular line. Such equations are all equivalent, even though they differ in
appearance. Here is an example of a pair of equivalent linear equations:
A. Y = <->0.5X + 4,
B. X – 2Y = <->8.
The equations appear different, but in this example the second equation was constructed by
multiplying both sides of equation A by <->2 and then adding X to both sides. In short, these
equations boil down to the same relationship between Y and X. Such a system is called
dependent because one of the equations can be found simply by rewriting the other equation
using the rules of algebra. These two equations represent only one distinct relationship
between the variables X and Y.
Because a linear equation can be written in different ways, it is possible to encounter a
system of two linear equations that turns out in reality to be two different ways of writing the
same equation! When we graph the two equations, they will appear as one and the same line
as in Figure 3.11.
{INSERT FIG. 3.11}
What happens algebraically if you try to use the substitution method on a dependent
system? Let’s try to solve the system above using the substitution method and find out.
Example 12
A Dependent System
Try to solve the system:
A. Y = <->0.5X + 4,
B. X – 2Y = <->8.
Chapter 3
Systems of Linear Equations and Inequalities
22
Solution
The top equation is already in slope-intercept form with the variable Y isolated, so we can
substitute (<->0.5X + 4) in the place of Y in the bottom equation to get
X – 2(<->0.5X + 4) = <->8.
Expanding the parentheses by the distributive rule gives us the equation,
X – 2(<->0.5)X – 8 = <->8,
which reduces to
X – X – 8 = <->8
and still further to
<->8 = <->8.
The bottom equation says that <->8 must equal <->8, which is always true no matter what
values of X and Y we might have in mind. The X and Y have fallen out of that equation. The
emergence of an equation that is always true is the algebraic signal that our original system of
equations was dependent. Therefore the solution set is the set of all (X, Y) pairs that satisfy the
first equation,
Y = <->0.5X + 4.
SUMMARY
This section introduced an important new type of mathematical problem: solving for more
than one variable at once when you have more than one equation relating the variables. This is
called solving a system of equations. The equilibrium price and quantity in a market is a
problem that requires solving a system of two equations. The problem of determining the
effectiveness of advertising, the efficiency of an employee, and the “prices” of buildings and
land can often be formulated in terms of a system of equations.
The technique for solving a system of two equations and two variables can be either
graphical or algebraic. Graphically, the solution to the system of equations is the point where
the linear graphs of the equations intersect. A graphical solution is at best an eyeballed guess.
The a more precise solution can be calculated algebraically by the substitution method in
three steps:
1. Isolate one variable in one equation.
2..Substitute the formula for that variable into the other equation and solve that
equation for the second variable.
3. Substitute that value for the second variable into the equation for the first variable
and solve to get a value for the first variable.
This method works almost all the time. There are two special cases where the substitution
method runs into trouble, and in both cases, it is because there is no unique solution to the
system of equations. The first, the inconsistent system, occurs when the two equations have
the same slope but different intercepts, meaning that they never intersect at all. The algebraic
signal of that problem is that the substitution method results in an equation that is always
false. The other problem, the dependent system, occurs when the two equations are just
different ways to write one equation. Then the solution is the entire line of the single equation.
This is indicated when the substitution method results in an equation that is always true.
Chapter 3
Systems of Linear Equations and Inequalities
23
EXERCISES
Skill Builders
For each of the following systems of equations, sketch a graph of the equations, and estimate
the solution by looking for the point of intersection. Then solve the system of equations
algebraically using the substitution method.
1. Q + P = 100,
Q = 0.75P – 1.5
2. Y = 4X + 10,
Y = 50 – 6X
3. 3X + 4Y = 270,
5X + 2Y = 100
4. 0.44A + 1.07B = 29.51,
0.80A + .55B = 24.35
2 X + Y = 13,
 X + 2Y = 11
 A + B = 17,
6. 
2A − B = 7
 3P − 4Q = 8,
7. 
 P + 2Q = 3.5
 X + Y = −2,
8. 
 X −Y = 6
 R = S + 150,
9. 
 R + S = 200
2 X + 3Y = 30,
10. 
 8 X − Y = 16
X Y
=1
 +
11.  30 20
 Y = X + 5
5. 
0.1X + 0.3Y = 11,
 0.2 X + 0.5Y = 19
12. 
Solve the following systems of equations using the substitution method, and check your answer
by substituting the solution into both of the original equations.
13. 3X – 6Y = 9,
2X + Y = 11
14. P = 200 – 5Q,
P = 60 + 2Q
15. <->5A + 8B = <->10,
15A + 2B = 160
16. Y – 9 = 2(X – 5,
Y – 1 = 3(X – 2)
Chapter 3
Systems of Linear Equations and Inequalities
24
7 X + 3Y = 29,
 X +Y = 7
 X + 2Y = 8.5,
18. 
 2 X − Y = 4.5
5 X − 6Y = 10,
19. 
 3 X + Y = 52
 X + Y = 12,
20. 
 X − 2Y = 18
5 X − 3Y = 23,
21. 
 4 X − Y = 24
V = 80 − 2U ,
22. 
V = 2U + 40
 4 P + T = 3,
23. 
3P − 0.5T = 6
Y − 3 = 2( X − 4),
24. 
 Y − 7 = 3( X − 5)
17. 
Use the substitution method to find out if the following systems of equations are either
inconsistent or dependent.
25. 3X + 5Y = 20
45Y = <->27X + 160
26. <->2X + 5Y = 40
Y = 0.4X + 8
 C − 2 D = 40,
C − 3 = 2( D + 18) + 1
2 X + Y = 15,
28. 
Y = 20 − 2 X
3 X − 2Y = 8,
29. 
Y = 1.5 X − 4
12 X + 16Y = 120,
30. 
 Y = 9.5 X − 0.75
27. 
Applications
31. Temperature. The centigrade, or Celsius, scale of temperature sets 0 degrees to be
the freezing point of water and 100 degrees to be the boiling point at sea-level pressure. The
Fahrenheit measure (F) is related to the centigrade measure (C) by the equation F = 32 + 1.8C.
Find the temperature at which the measurement in degrees centigrade is the same as the
temperature measured in degrees Fahrenheit. (Hint: What equation says that the temperatures
as measured by each scale are the same?)
Chapter 3
Systems of Linear Equations and Inequalities
25
32. Demand and supply. The quantity (Q) of pumpkins that would be demanded by
consumers at Halloween time in a particular town is estimated to depend on the price (P) of a
pumpkin by the equation Q = 25,000 – 2000P. Farmers in the region are not willing to plant
pumpkins unless they can expect to earn a profit on them, but if they foresee a high price of
pumpkins they will plant more acres of pumpkins and produce more. The farmers’ decisions
to produce a supply of pumpkins is captured by the equation Q = <->10,000 + 10,000P.
A. Draw a graph of the supply and demand “curves” in the pumpkin market.
B. At what price will the quantity demanded by the consumers be equal to the amount that the
farmers would be willing to supply?
33. Estimating productivity. The company Crystal Giftware produces two types of
product. The Sunview is a decorated disk of glass that hangs in a window frame. The Lamp
view is a prism of decorated rectangular glass panels that are soldered together and surround a
small light bulb that illuminates the design. Manufacturing employees work a 40-hour week.
The Marketing Department gives them a production request each week, detailing the number
of Sunviews and Lampviews that need to be produced to keep up with current and expected
demand. Production records indicate that when 6 employees were working, in 1 week they
produced 160 Sunviews and 80 Lampviews. In another week, they produced 70 Sunviews and
105 Lampviews. How many labor hours does it take to produce 1 Sunview? One Lampview?
34. Lakefront real estate. Along Lake Michigan in Illinois and Wisconsin, real estate
sells at a premium if it is right on the lake. For property that has no houses on it, it seems
reasonable to assume that the two most important factors in the cost of the lot are its acreage
and its frontage on the lake. A real estate agent has just provided you with data on the costs at
which several properties sold recently in the area where you want to invest in some land.
Property Acreage Frontage on
(A,
lake (F, feet)
Cost ($)
acres)
X
6.3
300
115,000
Y
24.0
500
300,000
Z
6.1
550
165,000
The real estate agent is offering you a small lot that has 2.0 acres and 150 feet of lake
frontage, for a cost of $40,000. Is this a good value?
A. The linear model says cost = cA + dF, where c is the price per acre and d is the price per
foot of lake frontage. Using the linear model, estimate the individual prices on acreage
and on frontage from the data above for properties X and Y.
B. Test your model by seeing how well it predicts the cost of property Z.
C. Based on the stated cost, should you buy the lot with 2.0 acres and 150 feet of frontage if it
is for sale at $40,000?
35. Advertising response. Eva markets her new cookbook, Root Vegetable Dishes from
Denmark, by taking out small advertisements in two magazines, Veggie Life and Vegetarian
Times. The ads in VL cost $100 each, and the ads in VT cost $175 each. People who want to
buy the book call a toll-free number and place the order with a “fulfillment service” firm that
takes their credit card number and ships the cookbook.
For 2 months, Eva has placed
advertisements in both magazines, more in VT
than in VL because VT has the larger
circulation. The number of ads and the
resulting number of orders are shown in the
table.
Month of
Ads
January
February
Veggie
Life
(number
of ads)
1
1
Vegetarian
Times (number.
of ads)
Book Orders
2
2
170
174
A. Eva wants to determine how many
responses are coming from each magazine so she can see how much response she is getting
per dollar of advertising spent at each place. Does the information in this table help her? If it
Chapter 3
Systems of Linear Equations and Inequalities
26
does, what advice can you give about the best place to advertise? If it doesn’t help her, explain
why not.
B. In March, Eva altered her advertising strategy and took out 2 ads in VL and only 1 in VT.
The result was 160 orders. Using this information with the February data, what do you
calculate to be the numbers of orders per ad in VL and in VT?
C. Which magazine had the higher ratio of sales per advertisement to cost per advertisement?
36. Diet planning. Suppose that the only items left in the college cafeteria are enchiladas
and pecan pie. The dietary ingredients of these foods and minimum requirements are as listed
in the table below. What combination of these foods would exactly fulfill an 18-year-old
female’s daily requirements for iron and vitamin A? (Hint: Your answer may seem odd, even
though your calculations are correct. Check your solution.)
Enchilada
Minimum daily
Dietary ingredient
(1 serving, 0.5 lb.)
Pecan Pie (1/6 pie)
requirement
Iron (mg)
3.3 mg/serving
4.6 mg/slice
15 mg.
Vitamin A (IUs)
2720 IU/serving
220 IU/slice
800 IU
37. Library usage. Ever since St. Catherine’s College began their new M.A. in Education
program, the usage of books in the library seems to have increased. However, the library does
not ask for a student’s status (graduate or undergraduate) when books are checked out, so it is
hard to say if the increased usage is due to the graduate students. The head librarian at St.
Catherine’s had the idea to compare usage at her institution to the library usage at St. Mary’s
College across town. St. Mary’s has a larger enrollment, but they have an even larger
percentage of graduate students due to their M.A. programmes in education, psychology, and
management. However, the librarian believes that the library usage per student of each type,
graduate and undergraduate, will be the same in each college.
The enrollment figures and library usage for the two colleges are given at right.
Undergraduate Graduate Library books
College
students
students
used per year
St.
1,100
100
33,500
Catherine’s
St. Mary’s
1,800
400
74,000
A. What does this information tell you about the number of books used per year by the typical
undergraduate vs. that of the typical graduate student?
B. Should St. Catherine’s consider charging a higher library fee to the graduate students?
38. A light money box. At the county fair, adult admission cost $4 and children’s
admission was $1.50. The turnstile at the entry counted a total of 400 people attending the fair,
but the booth attendants did not issue separate paper tickets for adults and children, as is often
done elsewhere.
At the end of the day, there was $900 in the money box. That greatly annoyed the county
commissioner, who thought that one of the booth attendants must have been taking money out
of the box. The commissioner was overheard to say, “There should have been a lot more adult
fees. I was at the fair most of the day, and I would say there were 3 kids for every 2 adults,
maybe 2 kids per adult at the most, but not more.”
Determine if the commissioner is right about the possible theft of fair revenues and decide
whether an investigation of the employees is warranted. (Hint: Does the number of adults and
children implied by the revenue data fit the commissioner’s estimates of the ratio of kids to
adults?)
39. Travel speeds. An aircraft flying between Des Moines and St. Louis covers a
distance of 200 miles each way. On the trip toward St. Louis, the aircraft faces a headwind
and takes 2 hours to make the trip. Returning to Des Moines, the trip takes only 1.5 hours.
What was the aircraft’s airspeed, and what was the speed of the wind? (Hint: The ground
Chapter 3
Systems of Linear Equations and Inequalities
27
speed (known distance/time) is equal to the airspeed minus the wind speed going into the
wind, but it is airspeed plus wind speed going with the wind.)
40. Canoe trip. Scott and Eva paddled their canoe on the Sudbury River one afternoon to
look at birds and wildlife. They started upriver and turned around after 45 minutes. The return
trip took 30 minutes, and as they passed under Lee’s Bridge, Scott noticed that the current was
flowing at about 8 inches per second.
How many miles did they paddle? (Hint: Convert the current speed to miles per hour and
look for two equations in two unknowns that say, “timespeed = distance.”)
41. Blend design. Two types of coffee beans, Brazilian and Ghanaian, are used to make a
blend of ground and roasted coffee. Suppose that Brazilian beans cost $0.80 per kilogram and
Ghanaian beans cost $0.45. What weight of Brazilian beans and what weight of Ghanaian
beans should be used in 1 kilogram of the blend to make the overall cost of the blend equal to
$0.60 per kilogram? (Hint: There are two equations in this problem. One describes the cost of
the blend, and the other describes the weight of the blend.)
42. Employee compensation plan. Mary and Tabatha are partners in a small consulting
firm, Employee Development Associates (EDA), that offers a seminar called “Exemplary
Customer Service” to business firms. When they started their company, both Mary and
Tabatha spent most of their time on marketing the seminar to the training directors of potential
client firms. After several months, they had developed some regular customers and then
started to spend more of their time on teaching the seminars. About 40% of their revenue goes
to cover office expenses, and the other 60% can be divided between the owners.
Mary is now expecting her first child. She wants to cut back her hours of work after the
baby comes, and she might withdraw altogether from EDA for a while. Tabatha will continue
to work full-time. This raises the question of how to divide the 60% of their revenues between
them, particularly because of all of the work that Mary put in on marketing to develop clients
who are now repeat customers and who will generate teaching revenues for EDA even while
Mary is on leave.
Their seminar involves 10 hours of instruction and usually has 15 participants for whom
the client firm pays a total of $800 per seminar. From past experience, Mary and Tabatha
agree that they have had to spend about 40 hours in sales efforts to land a contract for a course.
They want their compensation system to give the same rate of hourly pay to the marketing and
the teaching activities.
What percentage of the $800 seminar fee should go to the person who teaches it, and what
percentage should go to the person who landed the contract assuming:
A. Tabatha teaches the seminar and Mary landed the contract?
B. Tabatha teaches the seminar and both partners contributed equally to landing the
contract?
43. Compensation design. Oakley Industries currently pays its 20 salespeople a flat
salary of $2,000 per month rather than a commission on the sales that they make. As a result
of this policy, the company has been losing its good salespeople and attracting a lot of
mediocre ones. The sales manager is thinking about switching to a two-option commission
system that has the same total cost as before because he thinks that’s a way to compensate
both types of salesperson fairly.
Salespeople who choose the first option would get a $750 monthly salary plus a
commission equal to X% of their sales. This would be attractive to the ordinary salespeople,
who tend to sell about $10,000 of product per month. The salespeople who choose the second
option would earn no monthly salary but get Y percent of their sales. This policy has to be
more attractive to the hot salespeople, the type who can sell $25,000 per month. The sales
manager knows that not everyone can be a hot salesperson, but he does think that he can
recruit about 12 ordinary salespeople and 8 hot ones.
A. What commission rates X and Y (expressed as a percentage of sales) should he offer under
each option so that his total monthly sales person compensation is the same as it is now, and
yet enables a strong salesperson to earn $3,000 per month?
Chapter 3
Systems of Linear Equations and Inequalities
28
B. Under this arrangement will an ordinary salesperson who takes the first option earn more or
less than they are earning per month under the present system? Would an ordinary salesperson
rather take the second option? Would a hot salesperson prefer the first option?
Graphing Calculator Exploration
All of the above exercises can be solved algebraically and then checked using a graphing
calculator. But the following question requires repeated solution of the problem under
different assumptions. This will be easiest to solve using a graphing calculator that can
produce a solution after each change of parameters.
44. Criminal investigation. A bystander at the time of a bank robbery thought she saw
the thieves jump into a Royal Taxicab in making their escape, but she could not be sure of the
name “Royal.” Royal, which operates within the limits of a city 75 miles away by interstate
highway, uses only a specific brand of car whose gas mileage on highway and city roads is 35
and 21 miles per gallon (mpg), respectively.
One taxi was of unknown location at the time of the robbery. Royal Taxi records show
that this car’s tank had been filled that morning. By evening, the 12-gal tank registered 1/4
full, and the mileage was 250 more than what Royal’s fleet manager had recorded for that car
the day before. In similar legal cases where a car brand’s gasoline mileage data was used, the
courts allowed an error of ±5 mpg for highway and ±4 mpg for city driving. Might the taxi
driver be innocent? (Hint: You’ll have to solve this problem 4 times, once for each
combination of the extreme values for the mileage data.)
Group Exercise
45. Inconsistent systems of more than two equations. Crystal Giftware Corp. produces
two products, the Sunview and the Lampview. Their 6 manufacturing employees each work a
regular 40-hour week, unless they are absent due to family responsibilities or illness. Suppose
that the production data indicate the following.
Week
Regular hours
worked
Overtime hours Lampviews
worked
produced
Sunviews
produced
1
240
0
80
150
2
220
0
95
100
3
240
20
95
150
4
235
10
90
140
A. Each member of your group should pick a different pair of weeks. Use the data for that
pair of weeks to calculate how many labor hours are needed per Sunview and per Lampview.
Do you get a solution?
B. Compare all group members’ answers from part A. Do you get the same answers or
different ones? If you get different answers, then the system of all four equations, taken
together, can be called inconsistent. Think of a few reasons why you might be getting different
answers. (Think about the real production process that is summarized in the table of data
given here.)
C. Having only the data in this table, what do you recommend as a “best” estimate of the
number of labor hours that go into each Sunview and each Lampview?
D. In part C above, you answered a statistical question. We will not see much statistics in
this algebra course, but it is useful to know that statistics is the topic of mathematical study
where precisely this type of issue is addressed. Using algebra, you will always get an answer
to the question about labor hours for two products when you have two independent
observations. The answer comes straight from the technique of solving a consistent,
independent system of equations.
However, when you have more than two observations and still only two variables,
mathematically you have an “inconsistent system.” That is, there typically are no two labor-
Chapter 3
Systems of Linear Equations and Inequalities
29
hour coefficients that can solve all of the (here four) equations simultaneously. So you have to
settle for an estimate of the labor-hour coefficients that is a best approximation to the true
value. Statistics offers ways to think about what is a “best” approximation and offers
techniques to calculate that approximation.
Discuss these points in your group, and make sure that everyone understands (1) the
“problem” created by having more equations than variables, (2) the meaning of an
“inconsistent system” of equations, and (3) what the field of statistics has to offer toward a
solution of this problem.
3.2 Systems of Linear Inequalities
Learning Objectives
Graph a linear inequality.
Write an inequality given a verbal description.
Translate a table of data into a system of linear inequalities.
Find the solution set of a system of inequalities graphically.
Perform algebraic operations on, and invert, an inequality.
Spotlight on Dietary Planning
In the Spotlight that introduced the previous section, we looked at the problem of finding
a menu that fulfills specific dietary requirements. Our example was of a diet composed
entirely of spinach and potatoes. (The algebraic technique of that section works best with
systems of only two variables, so we could use only two foods in that problem.) By solving a
system of two equations in two unknowns, we found a menu that would fulfill an 18-year-old
female’s requirements for 60 milligrams (mg) of vitamin C daily and 15 mg. of iron.
{INSERT Photo 3.2}
The fact is, however, that there are many more than two dietary requirements that a menu
must fulfill. There is a minimum requirement for calories, protein, and various other vitamins
and minerals. A diet cannot be designed to serve exactly the minimum requirements of all
these. The diet must provide at least the minimum requirement. And for some ingredients, like
sodium and calorie content, the diet may have to fit within a maximum allowance. That is,
the diet would have to supply at most a certain amount.
In the previous section of this chapter, we studied the solution to a system of linear
equations. In this section, we will study problems that are expressed not as linear equations
but as linear inequalities.
A linear equation, such as Y = 3X + 6, is graphed as a line in the X-Y plane. An inequality
is a relation between two variables in which the “=” sign of a linear equation is replaced by
one of the inequality symbols: < (“less than”), > (“greater than”), ≤ (“less than or equal to”), or
≥ (“greater than or equal to”).
So, given a linear equation in the variables X and Y, such as
X + 3Y = 12,
the corresponding linear inequality could be any of the following:
X + 3Y < 12,
X + 3Y ≤ 12,
X + 3Y > 12,
or
Chapter 3
Systems of Linear Equations and Inequalities
30
X + 3Y ≥ 12.
Inequalities that use the symbols “<” or “>” are called strict inequalities. They say that a
quantity must be strictly less than or strictly greater than another quantity. The corresponding
inequalities using the symbols “≤” and “≥” are said to be not strict, because they allow the two
quantities to be equal.
In addition to the application of linear inequalities in dietary planning, there are many
important applications in management science, where problems of production and operations
management are often expressed in terms of limitations on managers’ decision variables.
Systems of linear inequalities have especially important applications in the management of
factories in which several products are manufactured. In those applications, the inequalities
define the limitations under which a production system must be operated.
SOLUTION SETS FOR LINEAR INEQUALITIES
The previous sections of this chapter have shown that the solution to a linear equation in two
variables is a line, and the solution to a system of two linear equations in two variables is
(usually) the single point where the two lines intersect.
Single inequalities and systems of inequalities, in contrast, will usually have many
solutions, comprising a region in the X-Y plane. It will be our task in this section to describe
the solution set either algebraically or graphically. The solution to one linear inequality is a
half-plane in the X-Y coordinate system that is specified by a boundary line, as the following
example shows.
Example 1
Solution Set of a Linear Inequality
Find the solution set of Y > 3X + 6 graphically.
Solution
The solution procedure has two steps.
1. Graph the boundary of the given inequality. In our example, the boundary is the graph of
the equation Y = 3X + 6. If the inequality is a strict inequality (strictly < or >), as in this
example, draw the boundary as a dashed line. If it is not a strict inequality (either ≤ or ≥ ),
draw a solid line.
The equation Y = 3X + 6 is in slope-intercept form, so we immediately recognize that Y =
6 is the vertical intercept. The horizontal intercept comes from setting Y = 0 and solving, thus
we find X = <->2. Since Y > 3X + 6 is a strict inequality, we draw a dashed line between these
two points, as in Figure 3.12.
{INSERT Fig. 3.12}
2. Shade in the solution region. To determine which side of the boundary line is the solution,
pick any point in the plane but not on the boundary line as a test point and substitute the
coordinates of that point into the inequality.
•
If the resulting inequality is “true,” then the half-plane containing that point is the
solution.
•
If the inequality is “false,” then the other half-plane is the solution.
We’ll use (0, 0) as our test point here. Substitute X = 0 and Y = 0 into the inequality Y >
3X + 6:
0 > 3(0) + 6 ?
Chapter 3
Systems of Linear Equations and Inequalities
31
0 > 6; false.
So the half-plane for the solution is the side of the boundary that does not contain our test
point (0, 0). We shade the side of the boundary that represents the solution, as shown in Figure
3.13.
{INSERT FIG. 3.13
Make a Note
The points (x, y) in the two-dimensional coordinate plane that satisfy the
equality form a boundary line for the solution to the inequality. The
boundary of the inequality is graphed as a line in the plane. The points (x, y)
that satisfy the linear inequality all lie on one side of the boundary line.
INEQUALITIES AS “CONSTRAINTS”
In common English, a constraint is a boundary or limit of some kind. For example, a student’s
lifestyle may be constrained by the amount of pocket money they earn in a part-time job.
When people talk about constraints, they often use words like “limitation” or “allowance,”
and they may refer to a “ceiling” or “floor” value to describe the way the constraint affects the
range of choices available to them. When working with, and writing, inequalities, these
English words signal the type of constraint, meaning whether it is of the ">" or "<" variety.
We’ll see how these words are used in the examples that follow.
Think about This
For each of the following words, think of an example where the word is
used to describe an inequality. What is the direction of the inequality in your
example? If you think the word could imply either direction, give an
example of each for that word.
allowance
requirement
limitation
ceiling
floor
Maximum Allowance Constraints (<, ≤)
Example 2
Dietary Planning
Consider the problem of creating a single meal for a college student who eats only enchiladas
and pecan pie.
The following table shows the content of 1 serving of enchiladas (1/2 pound) and a onesixth portion of pecan pie in relation to minimum or maximum requirements.  (The maximum
requirements on calories and sodium content have been made up for this example.)
Dietary ingredients
Enchilada
Pecan pie (1/6 piece) Daily requirement
“Nutritive Value of Foods,” Home and Garden Bulletin, No. 72 (Washington, DC: US Government
Printing Office, 2002), , <http://www.nal.usda.gov/fnic/foodcomp/Data/HG72/hg72.html>
Chapter 3
Systems of Linear Equations and Inequalities
Calories
Calories
Protein (g)
Iron (mg)
Sodium (mg)
235
235
20
3.3
1332
32
575
575
7
4.6
305
1,500 min.
2,500 max.
46 min.
15 min.
4,000 max.
Each enchilada provides 235 calories, and each slice of pecan pie has 575 calories. The
student must decide how many servings of each to take. Write and plot an inequality that
expresses the following constraint,
The number of calories provided by E servings of enchiladas, plus the number of
calories provided by P slices of pecan pie, must not exceed 2,500.
Solution
A student who eats a number E of enchiladas will take in 235E calories. Likewise, by eating a
total of P slices of pecan pie they will take in 575P calories. The diet expresses a maximum
allowance on calories, an upper bound or limit. So we must write this English statement
mathematically as a formula (for the calories) that is required to be less than the number
2,500. This calls for the “<” inequality,
235E + 575P < 2,500 (calories).
The boundary equation for this inequality is 235E + 575P = 2,500. Its intercepts are
2, 500
2, 500
≈ 10.6 for E and
≈ 4.3 for P. The inequality is not strict, so we draw the line
235
575
solid, as if to “shade in” the (good) points on the line. As a test point, we use (0, 0) and see
that it results in the true inequality 0 < 2,500. Thus, the origin is part of the solution region to
be shaded in. (See Fig. 3.14.)
{INSERT FIG. 3.14}
By a similar logic you can construct a maximum allowance constraint for the sodium
content of the meal in the following question.
Practice
Each enchilada has 1,332 mg of sodium, and each slice of pecan pie has 304
mg of sodium. Assuming that the student should limit their sodium intake to
2,000 mg in this meal, write out an English sentence for the maximum
allowance on sodium, and then write it as an inequality.
Answer: 1332E + 304P < 2,000 (mg sodium)
Minimum Requirements Constraints (>, ≥)
In a dietary context, there may be minimum requirements for the intake of essential
vitamins, minerals, and calories. These minimum requirements constraints all appear as the
“>” or “>” variety.
Example 3
Dietary Planning
Suppose that in terms of protein, a diet must satisfy a constraint expressed in English as
The number of grams of protein provided by E servings of enchiladas, plus the number of
grams of protein provided by P slices of pecan pie, must be at least 46.
Chapter 3
Systems of Linear Equations and Inequalities
33
Food research shows that enchiladas typically have 20 grams of protein per serving, and
pecan pie has 7 grams of protein per slice. Write and plot an inequality that expresses the
protein constraint.
Solution
We formulate the protein requirement as a “greater than or equal to” inequality,
20E + 7P > 46 (grams of protein).
The inequality is not strict, so the boundary line is part of the solution. We therefore draw the
boundary as a solid line, as shown in Figure 3.15. The test point (0, 0) does not satisfy the
inequality, so the solution region is on the opposite side of the boundary line.
{INSERT FIG. 3.15}
Figure 3.15 shows that, in terms of protein, the daily diet should consist of at least a little
more than 6 slices of pecan pie along with no enchilada, or at least a little more than 2-1/2
servings of enchilada along with no pecan pie, or a minimum given by various combinations
in between.
Practice
Assume that a healthy diet requires at least 1,500 calories intake per day and
15 mg of iron. Each enchilada provides 3.3 mg iron, and each slice of pecan
pie has 4.6 mg iron. Write out an English sentence for the minimum
requirement on calories and another sentence for iron. Then write these
mathematically as inequalities.
Answers: 235E + 575P > 1,500 (calories) and 3.3E + 4.6P > 15 (mg. iron)
SOLUTION TO A SYSTEM OF LINEAR INEQUALITIES
We have seen that the solution to one inequality is a half-plane. What would be the solution to
a system of inequalities?
When we studied systems of linear equations, a solution was a point in the coordinate plane
that simultaneously satisfied each of the two equations. The solution to a system of linear
inequalities will be the set of all points that satisfy simultaneously each of the inequalities in
the system.
A system of three or more equations in two variables will usually have no solution at all
because three lines do not ordinarily intersect all in one point. In contrast, a system of
inequalities in two variables can have solutions even when there are more than two
inequalities in the system.
Example 4
Diet Planning with Many Constraints
Let’s consider the minimum requirements for calories, protein, and iron in the college
student’s diet problem. The table below reproduces a portion of the data shown in Example 1.
Dietary ingredients
Calories
Protein (g)
Iron (mg)
Enchilada
235
20
3.3
Pecan pie (1/6 piece)
575
7
4.6
Daily requirement
1,500 minimum
46 minimum
15 minimum
Chapter 3
Systems of Linear Equations and Inequalities
34
Solution
Taken together, the requirements in this table can be expressed as a system of three
inequalities. To those three, we should add the obvious constraints that neither the number of
enchiladas eaten per day (E) nor the number of slices of pecan pie eaten per day (P) can be a
negative number. The resulting system is the following:
A. 235E + 575P > 1,500 (calories),
B. 20E + 7P > 46 (grams of protein),
C. 3.3E + 4.6P > 15
(mg. iron),
D. E > 0 (enchiladas),
E. P > 0 (pecan pie slices).
To graph this system of inequalities, we will graph each inequality on the same coordinate
system, shading in the half-plane that satisfies the inequality. The solution to the system is the
set of points that satisfy all the inequalities simultaneously. This will be the region where all
the shading coincides. On the same axes, we graph each inequality individually, following the
steps described in Example 1:
1. Graph the boundary line, using a solid line for a non-strict inequality (). Label the line with
a word that describes the constraint.
2. Pick a test point, such as (0,0), and substitute those coordinates into the inequality to
determine which side of the boundary line is the solution region. Shade the correct region using a
distinctive pattern for each inequality, such as horizontal, vertical, or angled hash marks.
The nonnegativity constraints D and E are easy to graph. Their boundary lines are the
horizontal and vertical axes. Together, these constraints shade in all the points above the
horizontal axis and to the right of the vertical axis.
The solution to the system is the region where all shading coincides. In this case, the
solution is the part of Figure 3.16 that lies above and to the right of all the boundary lines.
Draw a darker line along the boundary of that solution region to identify it clearly.
{INSERT FIG. 3.16}
A system of inequalities can combine minimum requirement constraints and maximum
allowance constraints. In the example below, we consider a more complete system of
inequalities for the diet problem, involving mixed constraints.
Example 5
Diet Planning with a System of Mixed Constraints
This table shows a portion of the table that was introduced with Example 2. It shows three
minimum requirement constraints and two maximum allowance constraints. Write a set of
inequalities that expresses these dietary requirements and graph the solution set for this
system.
Dietary
Pecan pie
Daily
ingredients
Enchilada
(1/6 piece)
Requirement
Calories
235
575
1,500
minimum
Calories
235
575
2,500
maximum
Protein (g)
20
7
46 minimum
Iron (mg)
3.3
4.6
15 minimum
Sodium (mg)
1,332
305
4,000
Chapter 3
Systems of Linear Equations and Inequalities
35
maximum
Solution
We rewrite the information from this table in mathematical form as inequalities. The left side
of each inequality is a formula written in terms of the quantity of enchiladas E and the number
of pecan pie slices P eaten per day for the dietary ingredient (calories, protein, etc.). The right
side of each inequality is the requirement or allowance for that dietary ingredient. The
direction of the inequality (< or >) is determined by the type of constraint (minimum
requirement or maximum allowance).
A. 235E + 575P > 1,500 calories (minimum requirement),
B. 235E + 575P < 2,500 calories (maximum allowance),
C. 20E + 7P > 46 grams of protein (minute requirement),
D. 3.3E + 4.6P > 15 mg iron (minimum requirement),
E. 1332E + 304P < 4,000 mg sodium (maximum allowance).
The food must be consumed in positive quantities, so we add the nonnegativity constraints,
F. E > 0 enchiladas (nonnegativity),
G. P > 0 pecan pie slices (nonnegativity).
The graph in Figure 3.16 already has the minimum requirement inequalities A, C, and D.
To those, we will add B and E to determine the solution set of this more complex system. You
can set up these equations yourself by following the steps in the Practice box below.
Practice
1. Find the vertical and horizontal intercepts of the two remaining
inequalities,
B. 235E + 575P < 2,500,
E. 1,332E + 304P < 4,000.
2. Which kind of line, solid or dashed, should represent the boundaries of
the inequalities B and E?
3. Does the point (0,0) test “true” or “false” when substituted into each
inequality?
Answers: 1. B, (0,100/23), vertical and (500/47, 0), horizontal. E, (0, 250/19), vertical and
(1,000/333, 0), horizontal
2. Both should be represented by solid lines since they are not strict inequalities, i.e., each of
them is “less than or equal to.”
3. The point (0,0) tests as “true” in both cases. So the half-plane containing (0,0) in both cases
is the solution region.
Figure 3.17 shows the final graph of the system. The solution set is represented by the
region where all of the shading overlaps. The coordinates of any point in that region
correspond to a feasible combination of enchiladas and pecan pie, given our diet constraints.
{INSERT FIG. 3.17}
Practice
By examining Figure 3.17, determine whether each of the following diets
does or does not satisfy all dietary requirements.
A. 2.5 enchiladas and 4 slices of pie
Chapter 3
Systems of Linear Equations and Inequalities
36
B. 4 slices of pie and 2 enchiladas
C. 2 enchiladas and 3 slices of pie
D. 4 slices of pie and 1 enchilada
Answers: (A) no, (B) no, (C) yes, (D) too close to the boundary line to tell from the graph
More than Two Variables?
In reality, of course, a healthy diet must include more than just these two foods. But when a
system of inequalities has more than two variables, it is not possible to graph the solution set
in a plane. The solution set would be a region of three-dimensional space for a problem with
three variables, a region in four-dimensional space for a problem with four variables, and so
on. These kinds of solutions are much harder to visualize. However, our analysis of the twovariable case lays the groundwork for understanding problems that have any number of
variables.
ALGEBRAIC OPERATIONS ON INEQUALITIES
An inequality in two variables defines a half-plane in the coordinate system. What algebraic
operations on an inequality preserve its solution set, and which operations change it?
Addition Rule for Inequalities
Adding (or subtracting) any real number to both sides of an inequality does not change the
solution set. For example, it is true that 3 < 5, so it is true that 3 + 2 < 5 + 2, which is to say 5
< 7.
This rule can be applied to inequalities that represent a relationship between two
variables. As an example, consider Y > 3X + 4. Its solution is the region above the boundary
line Y = 3X + 4. Adding a number to both sides of this boundary-line equation does not change
the boundary line or alter the solution region. Thus, another inequality having the same
solution region is
Y + 10 > (3X + 4) + 10.
Multiplication Rule for Inequalities
Multiplying (or dividing) an inequality on both sides by a positive number does not change the
solution set. For example, it is true that 3 < 5, and it is likewise true that 2  3 < 25, meaning 6
< 10.
As applied to inequalities with variables, consider Y > 3X + 4 again. Multiplying a
positive number on both sides of the boundary-line equation does not change the boundary
line or alter the solution region. Thus, another inequality with the same solution region would
be
10Y > 10(3X + 4).
However, multiplying or dividing an inequality by a negative number has the effect of
reversing the direction of the inequality. We know that 3 < 5. But <->1  3 is not less than <>1  5. The inequality becomes reversed: <->3 > <->5.
As applied to inequalities with variables, this means that when we multiply both sides of
an inequality by a negative number, we must reverse the direction of the inequality to
preserve its solution set. For example,
Y > <->2X + 5
has the same solution set as:
<->0.5Y < X – 2.5.
Chapter 3
Systems of Linear Equations and Inequalities
37
Example 6
Inverting an Inequality
The inequality P > <->5Q + 200 shows how to determine the permissible values of P given a
number for Q. Invert that inequality, so that it shows how to find the permissible values of Q
given a number for P.
Solution
Starting with the original inequality,
P > <->5Q + 200,
subtract 200 from both sides. This addition/subtraction operation does not change the
direction of the inequality! The result is
P – 200 > <->5Q.
Now divide both sides of the inequality by <->5. This division by a negative number forces us
to reverse the direction of the inequality, changing ">" to "<."
200
P
–
< Q,
−5
−5
<->0.2P + 40 < Q.
Now turn around the inequality, so it reads with Q on the left side,
Q > <->0.2P + 40.
SUMMARY
An inequality describes a situation where one quantity is to be greater than, or greater than or
equal to, or less than, or less than or equal to, another quantity. We saw that linear inequalities
can express minimum requirements (as “>" or “≥” constraints) or maximum allowances or
resource limitations (as “<“ or “≤” constraints).
The solution to an inequality involving two variables is the set of all points in the plane
that satisfy the inequality. These points will all lie to one side of the boundary line of the
inequality, which is the graph of the equation that corresponds to the inequality. The solution
to one inequality is a half-plane.
The solution to a system of inequalities is the set of points in the coordinate plane that
satisfy all the inequalities simultaneously. This region of points corresponds to the intersection
of the half-planes that solve each of the individual inequalities, and it is bordered by the graphs
of their boundary lines.
Algebraic manipulation of inequalities is similar to the solution of equations. The rules for
addition and the rule for multiplication by positive numbers are the same as for equations.
However, one rule is different. When you multiply or divide an inequality by a negative
number, you must change > to < or vice versa, which is called reversing the direction of the
inequality.
We may turn around an inequality, for example changing X > 2Y to 2Y < X. This is not an
algebraic operation but simply a logical restatement of the inequality. Notice the difference
between reversing the direction (due to multiplying by a negative number) and turning around
(restating) an inequality.
Chapter 3
Systems of Linear Equations and Inequalities
38
EXERCISES
Skill Builders
In the X-Y coordinate system, graph the solution set to each the following linear inequalities
taken by itself.
1. Y < 3X + 10
2. Y > 14 – 2X
3. 3X + 5Y < 15
4. 3X + 5Y > 15
5. Y < 10
6. X > 5
7. 2X – Y > 8
8. –2X + 3Y < 20
9. Y – 6 > 2(X + 3)
10. 2Y < 3X + 5
11. Y  3 X + 10
12. Y  −3 X + 10
13. Y  −3 X + 10
14. Y + 2 X  6
15. Y + 2 X  6
16. 2  X  5 (This means “ 2  X and X  5 .”)
17. −4  Y  −1
18. 2 X + 3Y  12
19. 2 X + 3Y  6
20. 6  2 X + 3Y  12
Graph the solution set for each of the following systems of inequalities.
21. 3X + 7Y ≤ 150
5X + 4Y ≤ 100
3X + 8Y ≤ 120
X≥0
Y≥0
22. X + Y ≥ 90
5X + 12Y ≥ 600
4X + 6Y ≥ 240
X≥0
Y≥0
23. 4X + 5Y ≥ 360
2X + 4Y ≥ 200
5X + 9Y ≤ 400
X≥0
Y≥0
24. 3X + 5Y ≤ 400
3X + 5Y ≥ 200
2X + 10Y ≥ 350
Chapter 3
Systems of Linear Equations and Inequalities
39
Y≥X
X≥0
Y≥0
X  0,


Y  0,

25. 
 40 X + 50Y  300,
 10 X + 5Y  35
X  0,


Y  0,

 6 X + 8Y  480,
26. 
5 X + 4Y  100,

X  60,

Y  40

X  0,


Y  0,

27. 5 X + 2Y  20,
6 X + 5Y  60,

2 X + 5Y  40
X  0,


Y  0,

28.  X + Y  15,

Y  0.5 X + 2,

 X + 3Y  30
Applications
29. Production constraint. A wood shop manufactures tables and chairs. Assume that
every table requires 15 minutes on the lathe machine, and every chair requires 20 minutes.
Write an inequality that says, “The time used for lathing T tables plus the time used for lathing
C chairs must not exceed 10 hours each day.” Sketch the graph of this inequality.
30. Inventory constraint. A computer retailer sells computers and monitors. Every
computer in inventory. Write an inequality that says, “The total inventory space that can be
occupied by C computers plus the inventory space occupied by M monitors must not be more
than 500 square meters.” Sketch the graph of this inequality.
31. Diet planning. Assume that every kilogram of rice provides 10,000 calories to a diet,
and every kilo of beans provides 7,500 calories. Write an inequality that says a diet of rice and
beans together must provide a minimum requirement of 2,000 calories per day. Sketch the
graph of this inequality.
32. Blending coffee. The bitterness of a blend of coffee is a weighted average of the
bitterness (as judged by expert tasters) of the types of coffee that are mixed in the blend.
Suppose that the bitterness index of coffee type A is 3 and that of coffee type B is 7. Write an
inequality that says the bitterness of a blend of A and B must not exceed the bitterness index
Chapter 3
Systems of Linear Equations and Inequalities
40
value of 5.5. Sketch the graph of this inequality. (Hint: As variables, use the percentage of
each type of coffee in the blend, and remember that those two percentages must sum to 100%.)
33. Budget constraint. Alice has a monthly budget of $40 for recreation. Suppose that
she goes to a small rural college where the only kind of fun is going to a movie or eating out.
A movie costs $5, and a dinner out costs $10. Write an inequality that shows what
combinations of movies and dinners out per month she can afford and sketch its graph.
34. College admissions. The Admissions Office at State University has a simple rule for
admitting applicants, which is based on the applicant’s high school grade-point average (GPA)
and their score on the SAT exam. A student will be admitted if they satisfy any of the
following three conditions: (1) their GPA is at least 3.5, or (2) their total SAT score
(maximum 1,600) is at least 1,200, or (3) the sum of their SAT score plus 200 times their GPA
is at least 1,750. Prepare a graph that shows which students will be admitted and which will
not.
35. Loan screening. A local bank gives out small business loans after evaluating the loan
applicant’s business plan and personal assets. The loan officer rates the business plan on a 010 scale and likewise rates the person’s assets on a 0-10 scale. The loan officer puts twice as
much importance on the quality of the business plan as on the personal assets, so the officer
forms a total score for the loan application by adding the assets rating to 2 times the business
plan rating. A loan application that gets a total rating of 21 or higher will be funded. Write an
inequality to describe the banker’s criterion for awarding small business loans and sketch a
graph to show which applications will get funded and which ones will not.
36. Health care screening. A doctor can make a good guess about the cause of a child’s
infection by taking the child’s temperature (T) and white blood cell count (W). A normal
temperature is 37 to 37.5oC. A normal white cell count is 4,000 to 8,000 cells per cubic
millimeter of blood. A bacterial infection is more likely than a viral infection if T > 39oC and
W > 15,000 cells per mm3.
Sketch a graph of the regions in the T-W plane that correspond to normal health, viral
infection, and bacterial infection. For the domain of temperature on the horizontal axis, use
values from 37oC (98.6oF) to 41oC (105.8oF).
37. Phase diagrams. Whether a chemical will be in a solid, liquid, or gaseous state
depends on the temperature and pressure of its environment. The graph that shows which
combinations of temperature (T) and pressure (P) results in which state of matter is called a
phase diagram for that chemical. The phase diagram is commonly plotted on a coordinate
system with T on the horizontal axis and P on the vertical axis. Three distinct regions in the TP plane correspond to the states of solid, liquid, and gas.
As an example, carbon dioxide (CO2) is a gas at room temperature (25oC) and sea level (1
atmosphere pressure). But if its temperature is decreased or its pressure is increased, it can
become a solid (dry ice) or exist in liquid form. In the case of carbon dioxide, the temperature
at which it becomes a gas is, to a fair approximation, linearly related to the pressure.
Specifically, carbon dioxide becomes a gas when T > 5.35  P – 83.4, with T measured in
degrees centigrade and P measured in atmospheres.
A. Over a domain of temperatures from <->100oC to <->0oC, sketch part of the phase diagram
for CO2. That is, graph the region on the T-P coordinate grid where carbon dioxide is in
gaseous form. (Hint: To draw it on a grid with P on the vertical axis, you may want to invert
the inequality so that it reads P < . . . .)
B. Carbon dioxide exists in liquid form only at pressures above 5 atmospheres. At those
pressures, it will be a solid, rather than liquid, only if T < 0.8P – 60. Add this inequality to the
previous one to complete your phase diagram for CO 2, and label the regions that correspond to
solid, liquid, and gaseous states.
 Pressure can also be measured in pounds per square inch. Atmospheric pressure at sea level
is 14.7 lb./sq. in. This is defined to be a pressure of “1 atmosphere.”
Chapter 3
Systems of Linear Equations and Inequalities
41
C. At the ordinary sea-level pressure of 1 atmosphere, what temperature is required to form
dry ice?
3.3 Characteristics of a System of Linear Inequalities
Learning Objectives
Determine the amount of slack in a constraint and evaluate its meaning.
Determine when a constraint is redundant.
Recognize when a system of inequalities is inconsistent.
Understand the reality behind unbounded solution sets.
Spotlight on Slack Time in the Admissions Office
“This is extraordinary,” said Joanne Fennell as she returned from the campus post office one
morning. “It looks like this week we’ll have 50 more applications for admission than last
week, and I’d say that about a third of them are from foreign students.”
“That’s both good news and bad news for me,” replied Corrine Early, the director of
Admissions. “The more applications the better, but it takes a lot of our staff’s time to review
them. I had to press our vice-president quite hard just to get three extra people last week to
help with admissions processing. I was hoping those employees would give our office a bit of
slack, but it now looks as though we’ll be back working at full throttle again.”
“Perhaps not,” said Joanne. “Each American application takes about 2 hours to review,
and each international application about 3.5 hours. I think those three people can handle the
extra load, but I will have to sit down and do a few calculations to make sure.”
{INSERT Photo 3.3}
SLACK CONSTRAINTS AT A POINT
Figure 3.18 shows two points within the solution region of a “minimum requirement” type of
linear inequality. At point A, which is on the boundary of the solution region, the inequality is
said to be binding, in other words, on the boundary line, the inequality is satisfied as an
equality. Point B lies in the interior of the solution region. At that point, the inequality is said
to be slack or “have slack.” This means that at point B, the right-hand side of the inequality is
more than the minimum required.
{INSERT Fig. 3.18}
The concept of a constraint being binding or slack at a point applies just as well to a
system of inequalities. If the point lies on the boundary of the solution set for the system, then
at least one constraint will be binding, meaning that that particular constraint has no slack. At
a point on the boundary of the solution where two constraint lines intersect, each of those
constraints will be binding, and all other constraints will be slack.
The analysis of slack constraints is very important in production and operations
management. We shall see that when a constraint is slack at a point, it means that the righthand side of the constraint can be increased or decreased, and yet the point will remain in the
solution region.
Chapter 3
Systems of Linear Equations and Inequalities
42
Example 1
Slack in the Calorie and Protein Constraints
Our solution to the diet problem was the region in the graph shown in Figure 3.19, including
the boundary lines.
{INSERT FIG. 3.19}
Notice that 2 enchiladas and 3 slices of pecan pie is a feasible daily menu in this example. At
this point, how much slack is there in the protein requirement, and how much in the calorie
allowance?
Solution
We look first at the mathematical statement of the protein requirement,
20E + 7P ≥ 46 grams of protein.
We substitute the values E = 2 and P = 3 into the left side of this constraint and find out how
much slack remains in the requirement,
20(2) + 7(3) ≥ 46 grams of protein,
61 ≥ 46.
Evidently, the slack in the protein constraint at the point (2, 3) is 61 – 46 = 15 grams of
protein. This diet provides 15 grams of protein more than the minimum requirement. Put
another way, the minimum requirement could be increased by 15 grams and this menu would
still be feasible.
Now, the calorie allowance says,
235E + 575P ≤ 2,500 calories.
We use the values E = 2 and P = 3 in this inequality to find out how much the left side remains
below the right side:
235(2) + 575(3) ≤ 2,500,
470 + 1725 ≤ 2,500,
2,195 ≤ 2,500.
The slack in the protein constraint is 2,500 – 2,195 = 305 calories, meaning that the diet
provides 305 calories less than the maximum allowance. Another way of looking at this is to
say that the calorie allowance could be reduced by 305 calories, to 2,195 per day, and the diet
of 2 enchiladas and 3 slices of pecan pie would still be feasible.
Example 2
Slack in a Manpower Constraint
Consider the limitation on the use of manpower in a wood shop to produce the quantities B
and T of bookcases and tables, respectively. If each bookcase requires 3 hours of labor and
each table uses 4 hours of labor, then the constraint that weekly production must not use more
than 80 labor-hours can be expressed by the inequality
3B+ 4T ≤ 80 labor-hours per week.
If this week’s production schedule calls for B = 12 bookcases and T = 10 tables, is there slack
in the constraint? If so, how much?
Solution
Chapter 3
Systems of Linear Equations and Inequalities
43
The total labor requirement will be 3(12) + 4(10) = 76 hours, which is 4 hours less than the
limitation in the constraint. So we would say that at the point (12, 0) in the B-T plane, the labor
constraint has 4 hours of slack. (See Fig. 3.20.)
The amount of slack in a constraint at any one point is the quantity by which we would
have to change the right-hand side of the constraint in order to make the constraint binding at
that point, as the following example demonstrates.
{INSERT FIG. 3.20}
Example 3
Slack in the Admissions Office
The opening Spotlight for this section introduced the example of a school that had received 50
more applications than they were expecting that week. International applications require 3.5
hours of staff time to process, but American applications require only 2 hours. The office had
received 3 new staff people, who each work 40-hour weeks. Can these people handle the
additional load of the new applications?
Solution
The constraint on staff time, stated in words, is that the total number of hours needed to
process the American applications plus the number of hours needed to process the foreign
applications cannot exceed the total of 120 hours of staff time available from these three
employees. If we write A for the number of American applications in the batch and F for the
number of foreign applications, the constraint can be stated mathematically as
2A + 3.5F ≤ 120.
The batch of 50 new applications appeared to be about one-third foreign and two-thirds
American, meaning about 502/3 ≈ 33 Americans and 501/3 ≈ 17 foreign. Will that leave any
slack in the staff-time constraint? Figure 3.21 shows a graph of this inequality together with
the point (A,F) = (33, 17), which represents the needed manpower. The point seems to be
outside the solution region, but it is too close to the boundary line to tell visually.
{INSERT FIG. 3.21}
Mathematically, we can determine precisely whether or not the point is in the solution set.
We find out by substituting F = 17 and A = 33 into the inequality and testing the truth of that
mathematical statement:
2(33) + 3.5(17)
≤ 120
66 + 59.5
≤ 120
125.5
≤ 120; False. It is not in the solution set.
The conclusion is that 33 American applications and 17 foreign applications would more
than use up the available time of the new staff. There will not be any slack personnel time in
the Admissions Office this week!
Practice
1. What is the amount of slack in the inequality
5X + 4Y < 100 (hours of machine time)
at the point (X,Y) = (9,12)?
2. What is the amount of slack at the point (10,13)? What does this tell you
about the slack in a constraint at a point that is outside the constraint’s
solution set?
Chapter 3
Systems of Linear Equations and Inequalities
44
Answers: (1) 7 hours of machine time. (2) The slack is negative at points outside the
solution set of that constraint.
REDUNDANT CONSTRAINTS
In the course of solving realistic problems, it is not unusual to discover a new limitation
constraint or a new requirement to be fulfilled. Adding the new constraint may change the set
of points in the solution. Or, as you will see in the next example, it may not.
When an inequality can be added to a system, or removed from the system, without
changing the solution set, that inequality is called redundant.
Example 4
Diet Problem
Returning to our diet problem, assume that a minimum daily requirement of 1,000 IUs of
vitamin A has been added to our list of constraints. Each enchilada provides 2,720 IUs of
vitamin A, and each slice of pecan pie provides 220 IUs.
Write an inequality that describes this constraint and decide if it’s redundant.
Solution
Express this constraint as 2,720E + 220P > 1,000 (IUs of vitamin A). The vertical and
horizontal intercepts of this expression give an idea of where this constraint lies in the
enchilada-pie plane. Set E = 0 and you see that the constraint can be met by
4.5 slices of pie. But setting P = 0 and solving for E tells us that
1,000
or about
220
1,000
or about 0.37 of an
2,720
enchilada alone will meet the constraint.
The vitamin A constraint will have a boundary line that passes between these two
intercepts, as shown in Figure 3.22. When this constraint is added to the set of minimum
requirements and maximum allowances that was graphed in section 3.2, we see that the
vitamin A constraint falls far away from the set of solution points.
{INSERT FIG. 3.22}
This tells us that adding or removing the vitamin A constraint from the other five constraints
does not change the solution. The vitamin A constraint is therefore redundant.
Another example is seen in the following system of inequalities.
Example 5
Identifying Redundant Constraints
Identify the redundant constraints, if any, in the following system:
1. 2X
+ 5Y
2. 7X + 4Y < 28,
3. 2X + 2Y < 12,
4. X ≥ 0,
5. Y ≥ 0.
< 15,
Chapter 3
Systems of Linear Equations and Inequalities
45
Solution
When we plot these on the same coordinate system, we get the graph shown in Figure 3.23.
{INSERT FIG. 3.23}
The solution set is the region in the lower-left portion of the graph, not including the boundary
lines because the inequalities are strict (<). Inequality 3 does not pass through this region, so
its removal from the system would not alter the solution set.
Another way to see this is that adding inequality 3 to the system does not restrict the
solution set any more than what the inequalities 1, 2, 4, and 5 together do. Inequality 3 is
redundant in this system.
In contrast, the attempt to remove any of the other constraints, 1, 2, 4, or 5 would alter the
region that represents the solution. If inequality 1 were removed, the region would then extend
up as far as the line for inequality 3. If inequality 2 were removed, the region would then
extend out to the right as far as the line for inequality 3. These facts show that inequalities 1
and 2 are not redundant within the entire system of inequalities.
INCONSISTENT SYSTEMS OF INEQUALITIES
In the diet example with pecan pie and enchiladas, the solution set was a small bounded region
in the plane. There are other types of solution sets for systems of linear inequalities, and
sometimes a solution may not exist at all.
The graph of the solution to the following system appears in Figure 3.24:
1. 2X
+ 5Y
≤ 15,
2. 7X
+ 4Y
≤ 28.
The figure also includes the inequalities X ≥ 0 and Y ≥ 0.
{INSERT FIG. 3.24}
Now consider a third inequality,
3.
2X
+ 2Y
≥ 12.
When the graphical solution of inequality (3) is added to Figure 3.24, there appear to be
two separate solution regions (Fig. 3.25). But the solution to a system of inequalities is the
region that satisfies all inequalities simultaneously. The appearance of two distinct regions
means that there is no point or region that satisfies inequalities 1, 2, and 3 simultaneously,
along with inequalities X ≥ 0 and Y ≥ 0.
{INSERT FIG. 3.25}
Such a system of inequalities is called inconsistent and has no solution.
UNBOUNDED SOLUTION SETS
A system of inequalities can have a solution that is infinitely large. If we consider only the
minimum dietary requirements in the diet problem, an infinitely large lunch would still meet
all of the constraints. This is evident in Figure 3.26 from all of the shaded space (solution set)
that goes infinitely far in the upper-right direction of the coordinate system. Only when we
imposed a maximum allowance constraint, such as the one on sodium or on calories, did the
solution set become a bounded region in the coordinate plane.
Chapter 3
Systems of Linear Equations and Inequalities
46
{INSERT FIG. 3.26}
Realistic problems in operations management always have bounded solution sets. For
example, suppose you are using linear inequalities to describe the production of tables and
chairs with limited amounts of time, labor, and raw materials. If you graph the inequalities and
come up with an unbounded solution set, it is an indication that there is a numerical mistake
somewhere in your system of inequalities or that some limitation constraint is missing entirely.
No production system can produce an infinite amount of a product.
For systems of inequalities that describe the requirements of a diet or of an animal feed,
the solution set often is unbounded. But that does not mean that the dietitian in a hospital,
school, or military base is going to prescribe a diet that gives each person an infinite amount of
carrots. In practical applications of the diet problem, the linear inequalities are used as a
backdrop for a problem of trying to find a diet that meets the requirements at lowest cost. In
the search for a lowest-cost menu, a dietitian avoids the unbounded part of the solution region
even though it is mathematically feasible as a diet.
LINEAR PROGRAMMING
The actual decisions faced by a production manager are made with some purpose or objective
in mind, such as to maximize profits or to minimize costs, under the constraints imposed by
the physical production system and other organizational requirements. The most important
application of linear inequalities, which we have studied in this section, is to define the set of
feasible choices in this type of decision problem. The technique that is often used to find the
maximum-profit point or minimum-cost point among the feasible points is called linear
programming.
The problem of maximizing a linear objective function subject to a collection of linear
inequality constraints is called a linear programme. There is an elegant solution to the
standard linear programming problem, which works for any number of variables and any
number of constraints when the inequalities have at least one solution. That solution procedure
is called the simplex algorithm. The mathematical ideas behind it are an application of matrix
algebra, which is studied in more advanced courses.
The simplex algorithm does more than just locate the values of the decision variables that
maximize or minimize the objective. In the process of calculating the solution, the simplex
algorithm provides vital information about the effects of changing the numerical values of the
constraints or of the decision variables. These insights provided by the simplex algorithm’s
solution go beyond what we can see in a simple graphical solution. In fact, mathematical
analysts use computers to calculate the complete solution to a linear programming problem in
any realistic application to profit maximization or cost minimization.
The subject of linear programming is so important to business decision making that the
topic rightly deserves to be a major part of a separate course. Business majors will encounter
the theory and practical applications of linear programming in a course on operations research
or on “quantitative methods” in business. Understanding the theory of linear inequalities, as
presented in this section, provides a basis for the study of linear programming.
SUMMARY
Systems of linear inequalities are a mathematical tool used in many important applications in
management and the social and health sciences, particularly in problems of maximizing profit
or minimizing cost. A system of linear inequalities is often used to describe the constraints
within which a person must make decisions. In this section, we followed an example where
linear inequalities define the set of menus (albeit a rather limited menu) that meet a whole
variety of constraints on a person’s diet.
When a solution to a system of linear inequalities exists, the solution set may be bounded
within a finite region of possibilities, as in our diet planning example (Fig. 3.19), or the
solution set may be unbounded (Fig. 3.26). Given any point in the solution region, it is
Chapter 3
Systems of Linear Equations and Inequalities
47
possible to calculate the amount of slack in a constraint, which tells us how much the righthand side of the constraint can be changed before the constraint becomes binding.
We noted that sometimes a constraint can be removed or added without altering the
solution to the other inequalities. Such a constraint is redundant within the system of
inequalities.
A system of linear inequalities is inconsistent if there is no point in the coordinate plane
that can satisfy all constraints simultaneously.
EXERCISES
Skill Builders
For each of the following systems of inequalities:
A. Graph the solution set.
B. Determine if the system is inconsistent.
C. Determine if the solution set is unbounded.
D. Identify the redundant constraints, if any.
[ED: Please take out tabs; not needed for comp]1.
5X + 4Y ≤ 100
3X + 8Y ≤ 120
X≥0
Y≥0
2.
X + Y ≥ 90
5X + 12Y ≥ 600
4X + 6Y ≥ 240
X≥0
Y≥0
3.
4X + 5Y ≥ 360
2X + 4Y ≥ 200
5X + 9Y ≤ 400
X≥0
Y≥0
X  40
Y  50
4. X + Y  100
X  10
Y  20
7 X + 4Y  280
10 X + 12Y  600
5. X + Y  60
X 0
Y 0
3X + 7Y ≤ 150
Chapter 3
Systems of Linear Equations and Inequalities
48
30 X + 20Y  1, 200
40 X + 50Y  2, 000
6.
Y  2 X + 10
X 0
Y 0
8 X + 7Y  280
X + 2Y  60
7.
2 X + 3Y  60
X 0
Y 0
8. In the following system of inequalities, which constraints, if any, are redundant, and which
form part of the boundary of the solution set?
A.
B.
C.
D.
E.
4X + 5Y < 360
2X + 4Y < 200
3X + 7Y > 300
5X + 4Y < 280
3X + 8Y > 240
9. In the following system of inequalities, which constraints, if any, are redundant, and which
form part of the boundary of the solution set? Pick one point at the intersection of two
boundary lines. Calculate the amount of slack in each nonredundant constraint at that point.
9 X + 11Y  495
2 X + 3Y  90
5 X + 6Y  150 (Note to compositor: this should have (A-E labels as above).
Y  15 + X
Y  6 X − 12
Applications
10. Publicity. The student association has allocated $600 toward publicity for the
upcoming Halloween Fest. Ads in the school newspaper cost $15 per column-inch. Color
posters can be printed and posted around campus for $1.25 each.
A. Write an equation for the publicity budget constraint and sketch its graph.
B. The Halloween Fest director thinks that to publicize the event properly, she will need to use
25 column-inches of advertising and distribute 80 posters. Is this plan feasible given her
budget constraint?
C. How much slack is there in the budget constraint if she follows the plan described in part
B?
11. Production constraints. Assume that every unit of product A requires 7 minutes on
the grinding machine and every unit of product B requires 12 minutes.
A. Write an inequality that says, “The time for grinding A units plus the time used for grinding
B units must not exceed 8 hours each day,” and sketch its graph.
B. How much slack is there in the grinding constraint if the current production schedule calls
for 18 units of product A and 25 units of product B to be manufactured today?
Chapter 3
Systems of Linear Equations and Inequalities
49
12. Inventory constraints. Assume that every unit of the radios in inventory uses 1.6
square feet. of space, and every unit of clocks uses 0.7 square feet.
A. Write an inequality that says, “The total inventory space that can be occupied by X units of
radios plus the inventory space occupied by Y units of the clocks must not be more than 500
square feet.”
B. How much slack remains in the inventory space constraint if the warehouse currently has
180 units of radios and 250 units of clocks?
13. Diet planning. Assume that every kilogram of dry rice provides 10,000 calories to a
diet, and every kilogram of dry beans provides 7,500 calories.
A. Write an inequality that says a diet of rice and beans together must provide a minimum
requirement of 2,000 calories per day.
B. Suppose that a person has already eaten 100 grams (0.10 kg) of dry rice and 80 grams (0.08
kg) of dry beans? Put a dot on your graph indicating this point. How much slack remains in the
calorie constraint at that point? What is the meaning of negative slack?
14. Dietary constraints. Suppose that a diet consists entirely of enchiladas and pecan pie.
In this problem, we consider the dietary constraints on fat. Each gram of fat, saturated or not,
contributes 11 calories to one’s diet. The table shows how much fat is contained in each
serving of enchiladas and pecan pie
Content of foods Enchilada
Pecan pie slice
(230 grams) (1/6 pie, 138 grams)
Fats (all kinds)
16 grams
32 grams
Saturated fats
7.7 grams
4.7 grams
Energy
235 calories 575 calories
A. Write inequalities that express the following 1996 dietary guidelines of the U.S. Dept. of
Health and Human Services.
•
No more than 30% of the calories in the daily diet should come from fats.
•
No more than 10% of the calories in the daily diet should come from saturated fats.
B. Are either of these constraints redundant when considered along with the constraints used
in the version of the diet problem shown in Figure 3.19?
15. Admissions screening. The Great Plains School of Journalism accepts students into
its master’s programme based on their verbal and analytical scores on the Graduate Record
Exam (GRE). The school considers verbal ability to be slightly more important than analytic
skills. Applicants are automatically admitted if the verbal score is at least 600, the analytical
score is at least 550, and the sum of the verbal and analytical scores is at least 1,230.
A. Sketch a graph of the regions in the verbal-analytic plane that correspond to automatic
admission.
B. On your diagram, mark the point corresponding to an applicant who has a verbal score of
590 and an analytic score of 650. Will this person be automatically admitted?
C. Continuing from part (b), how much error could there be in the applicant’s verbal score
before it would change the Admissions committee’s decision?
16. Admissions screening. The International Institute of Business admits students into its
MBA program based on their English-language proficiency as measured by the TOEFL exam
and their aptitude for business study as measured by the GMAT. They do not consider
undergraduate GPA because their students come from many countries with different grading
systems. Students at IIB are automatically admitted if their TOEFL is at least 575, their
GMAT is at least 550, and the sum of the TOEFL and GMAT is at least 1,200.
Chapter 3
Systems of Linear Equations and Inequalities
50
A. Sketch a graph of the regions in the TOEFL-GMAT plane that correspond to automatic
admission.
B. On your diagram, mark the point corresponding to an applicant who has a verbal score of
625 TOEFL and a GMAT score of 580. Will this person be automatically admitted?
C. Continuing from part (b), how much error could there be in the applicant’s TOEFL score
before it would change the Admissions committee’s decision?
17. Health screening. When a doctor evaluates a patient’s risk of heart disease, she
considers the patient’s age, gender, total cholesterol, and systolic blood pressure among other
factors. Write C for the total cholesterol (mg/dL) and B for the blood pressure (mm Hg). A
50-year-old man will increase his risk of heart disease in the next 10 years by 10 percentage
points if the measures of cholesterol and blood pressure satisfy the inequality 0.05C + 0.10B
> 30.6.
A. Sketch a graph of the regions in the C-B plane that correspond to less than 10% extra risk
and more than 10% extra risk heart attack. For the cholesterol measure use values between 140
and 330, and for the blood pressure measure use values between 100 and 180.
B. On your diagram, mark the point corresponding to the observation of a 50-year-old male
with a total cholesterol of 220 and a systolic blood pressure of 140. Has this person added
more than 10 percentage points to their chance of heart disease?
C. How much would the patient have to increase or decrease their blood pressure in order to
change the doctor’s assessment of the risk category that person falls into (that is, more than
10% extra risk or less than 10% extra)?
18. States of matter. Propane is used as a fuel for lamps, heaters, and stoves. At the sealevel atmospheric pressure (14.7 lbs. per square inch, or 1 atmosphere), propane changes from
liquid to gas at <->44oC. When the pressure is 17 atmospheres, propane becomes gaseous at a
temperature of 40oC.
A. Write an inequality that specifies the combinations of temperature and pressure at which
propane is in a gaseous state.
B. What amount of pressure is necessary to make propane liquid at the freezing point of water,
0oC?
19. Production constraints. Finney’s Wood Shop produces only tables and chairs.
Thomas Finney, the owner, has three resources used in production, namely, inventory space,
assembly labor, and painting labor. His shop is unionized, so the assembly workers do not do
painting and vice versa. The requirements of space and labor for one table and for one chair
are shown in the table below, as is the available quantity of each resource for the next week.
Unit production
Per
Per
Quantity
requirements
table
chair
available
Inventory space (sq. ft.)
16.00
2.00
1,200
Assembly labor (hrs./wk.)
0.75
0.60
120
Painting labor (hrs./wk.)
0.50
0.20
80
Finney wants to make plans for the production of tables and chairs next week. He already
has on hand orders for 40 tables and 100 chairs, which must be produced next week, or else
customers will be very unhappy.
A. Write five inequalities that define the constraints on Finney’s production plan for the next
week. Hint: Three inequalities will pertain to resource limitations, and two will pertain to
production requirements.
 K. M. Anderson et al. "An Updated Coronary Risk Profile," Circulation 83 (1991): 356-362.
Chapter 3
Systems of Linear Equations and Inequalities
51
B. The feasible combinations of tables and chairs constitute a region in the coordinate system
shown in Figure 3.27. Label each line in the diagram with the name of the appropriate
constraint and shade the solution region for each.
C. Give an example of a feasible quantity of tables and chairs for next week’s production.
{INSERT FIG. 3.27}
Analysis
20. Exercises 15 and 16 show that in an admissions process it is possible to describe the
“automatic admission” criteria by a system of linear inequalities.
A. In principle, can the criteria for “automatic rejection” be described by a system of linear
inequalities?
B. If you think it is possible, make up such criteria that are consistent with the data in Exercise
15.
3.4 Solving Systems of Linear Equations with More than Two Unknowns: Elimination
Method
Learning Objectives
Solve a system of linear equations using the elimination method.
Mathematically model a problem given its background and a table of
data.
Recognize algebraically when a system is either inconsistent or
dependent.
Spotlight on Decisions about Advertising
“We should have instructed the receptionist to ask callers how they heard about these new
programs,” said Donald Martinez, Assistant Director of Admissions. He looked over a report
of the number of inquiries that Metropolitan University in Chicago had received over the last
few months about its new graduate certificate programs in management. “Then we would
really know which of our advertisements are producing results and which are not worth the
money.”
Carol Haines, a business major who was working in the Admissions Office several hours
a week, overheard Mr. Martinez. She turned to him and asked, “How much advertising have
you been doing?”
“More than $10,000 per month. Why?”
“I actually meant to ask how many different magazines or newspapers you are advertising
in,” added Carol.
Mr. Martinez quickly rattled off a list of four of the most prominent papers in the Chicago
region. “I see,” said Carol. “How many months have you been collecting data about the
inquiries for your graduate certificate programs?”
“August, September, October, and November,” replied Mr. Martinez.
“I think I can tell you how many responses are coming from each ad,” said Carol with a
smile.
“You’re kidding. How can you do that if we didn’t ask the callers where they heard about
us?”
“Using a technique I learned in an algebra course a while back. If you can tell me how
many ads you placed in each source during each month, August to November, and you have
the total inquiry data for those months, it will take me about 15 minutes to set up the problem
on our office computer and solve it.”
Chapter 3
Systems of Linear Equations and Inequalities
52
“That would be great! I have a report on advertising costs and total inquiries right here. It
should contain the information you are asking for.” replied Mr. Martinez. He gave Carol the
following table.
{INSERT Photo 3.3}
Source:
Fortune
Chicago
Chicago
Magazine
Crain’s
Chicago
Business
Suburban
papers
Price/ad:
$3,500
$3,700
$2,300
$2,400
August ads
September
ads
October ads
November.
ads
1
0
0
1
1
1
0
1
1
0
0
1
Inquiries
received
3
2
Total
monthly
cost
$13,000
$10,800
2
2
$8,500
$10,600
300
440
520
445
In the previous sections of this chapter, we saw how to solve a system of two equations in
two unknowns. However, some problems in business and society require the simultaneous
solution of three or more equations.
As long as the equations are linear, there is a straightforward method to find a solution to
such a system. This technique is called the elimination method. We can use it to solve a
system of equations where there is a unique solution (analogous to a pair of crossed lines in
the two-variable case) and also to identify the unusual cases where there is no solution or an
infinite number of solutions.
To understand how Carol would solve Mr. Martinez’ problem, we will first look at the
elimination method in the simple context of two variables.
THE ELIMINATION METHOD WITH TWO VARIABLES
The elimination method works on a system of linear equations that has been written in general
form. Consider the following system of two equations in two unknowns:
A. 2X + 3Y = 10,
B. 4X – 8Y = <->15.
The objective of the elimination method is to reduce this system to an equivalent system in the
form
X = _____,
Y = _____,
where there are only numbers on the right-hand sides.
Another way of looking at this is to say that we want the first equation in the equivalent
system to have a coefficient of 1 on the X and 0 on the Y, and vice versa for the second
equation,
1X + 0Y = _____,
0X + 1Y = _____.
This technique is called the “elimination” method because the objective is to “eliminate” all
but one variable from each equation. The way to “eliminate” a variable is to reduce its
coefficient to zero following specific rules that do not change the solution to the system of
equations.
Chapter 3
Systems of Linear Equations and Inequalities
53
How does this method work in practice? The rules of algebra permit us to multiply an
equation, on both sides, by a nonzero constant. At first it may be a surprise, but the rule of
adding equal quantities to both sides of an equation also permits us to add one equation to
another equation—left side to left side and right side to right side—because both sides of the
one equation are equal.
The key to solving a system of equations by the elimination method is to choose a
multiplication factor for one equation in such a way that it will eliminate the coefficient of a
variable in another equation when the two equations are added.
Example 1
Solving a System of Two Equations in Two Variables
Using the elimination method, solve for X and Y in the following system:
A. 2X + 3Y = 10,
B. 4X – 8Y = <->15.
Solution
To eliminate the X variable from the bottom equation, we must add <->4X to the existing 4X,
which will result in 0X. Multiply equation (a) by –2,
<->2(2X + 3Y) = <->2(10),
resulting in the equivalent equation
<->4X – 6Y = <->20.
Now add this to equation (b), left side to left side and right side to right side, and call the
resulting equation B1,[comp: align on equal signs]
<->4X – 6Y = <->20,
4X – 8Y = <->15,
B1. 0X – 14Y = <->35.
Recall that our objective in this process is to create a system of equations that has only 0’s
and 1’s as coefficients. In particular, we want the bottom equation to have 0 as its X<−1
>coefficient and 1 as its Y-coefficient. We accomplish this by multiplying equation B1 by
:
14
−1(−14 )
−1
−35
(B1) 0X +
Y=
−14
14
14
= (B2) 0X + 1Y = 2.5.
Now we replace the original equation B by equation B2. The new system of equations has
the same solution as the original system of equations, but now it appears in a simpler form as
A. 2X + 3Y = 10
B2. 0X + 1Y = 2.5
If we were following the substitution method, at this point we would substitute the known
Y value, 2.5, into equation A and solve for X. However, the full-blown elimination method
must continue with two more steps. (You’ll see later in this section that the substitution
method would be very messy if we had more than two variables, yet the elimination method
would continue to be quite simple. Here, we are using the elimination method on a twovariable problem just to show how it works.)
The Y in equation A has a coefficient of 3. Our goal is a Y coefficient of 0, so we multiply
equation B2 by –3
Chapter 3
Systems of Linear Equations and Inequalities
54
<->3(b2)
+ (A)
0X – 3Y = <->7.5
2X + 3Y = 10
= (A1) 2X + 0Y = 2.5
To finish the elimination method, we must turn the coefficient of X in equation A1 into a
1
1. To do so, we multiply the entire equation A1, left and right, by ,
2
1
1
(A1)
(2X + 0Y)
2
2
=
= (A2)
= 1.25
1X + 0Y
1
(2.5)
2
The new system of equations, using A2 and B2, has the same solution as the original
system, but the solution is now immediately evident:
A2.
1X + 0Y = 1.25, X = 1.25,
which reveals that
B2.
0X + 1Y = 2.5, Y = 2.5.
We check the solution (X,Y) = (1.25, 2.5) by substituting these values into each equation
of the original system:
equation A:
2X + 3Y = 10,
2(1.25) + 3(2.5) = 10,
2.5 + 7.5 = 10;✓
equation B:
4X – 8Y = <->15,
4(1.25) – 8(2.5) = <->15,
5 – 20 = <->15.✓
Make a Note
The elimination method transforms the appearance of a system of linear
equations in a way that does not alter the solution to the system. In its final
form, the system has coefficients of 1 and 0, which enable us to “read” the
solution directly.
SOLVING EQUATIONS WITH THREE OR MORE VARIABLES
The following is an example of a system of three equations in three variables:
A.
X + 4Y + 2Z
= 29,
B.
X + 3Y + 3Z
= 35,
C.
3X + Y + 2Z
= 33.
Trying the Substitution Method on Three Variables
Before we apply the elimination method to this problem, let’s see why the substitution method
becomes difficult with three or more variables. The first step in the substitution method would
be to isolate one variable in one of the equations. The variable X is isolated in equation A, so
we isolate it by rewriting equation A as
X = 29 – 4Y – 2Z.
The next step in the substitution method would be to substitute this formula for X in the
other two equations, which then have only two variables each,
Chapter 3
Systems of Linear Equations and Inequalities
B1.
(29 – 4Y – 2Z) + 3Y + 3Z
= 35,
C1.
3(29 – 4Y – 2Z) + Y + 2Z
= 33.
55
This is now a system of two equations in two variables, Y and Z. We would have to simplify it
by collecting like terms,
B1.
<->4Y + 3Y – 2Z + 3Z
= 35 – 29,
C1.
3(<->4)Y + Y + 3(<->2)Z + 2Z
= 33 – 329,
and further simplifying to
B1.
<->1Y + 1Z
= 6,
C1.
<->11Y + <->4Z = <->54.
At this point, one could continue the substitution method, writing equation B1 as Z = 6 +
Y and substituting into equation C1 to get an equation in Y alone,
<->11Y + <->4(6 + Y)
= <->54,
<->11Y + <->24 – 4Y
= <->54,
<->15Y = <->30,
Y = 2.
With that value for Y in hand, the substitution method would continue by using that value
in equation (B1) to find Z = 8, and then using the known values for Y and Z in equation A to
find X = 5.
The first substitution step above created a rather messy system of two equations in two
unknowns, but we were still able to push forward with the substitution method. If the system
had had four equations in four variables, the first substitution step would have resulted in a
very messy system of three equations in three variables, which would be even more difficult to
pursue under the substitution method.
In contrast, the elimination method will be more straightforward to apply and can be
generalized much more easily to problems with four or more variables. We’ll now outline the
steps of that procedure.
Overview of the Elimination Method
The basic elements of the elimination method are the two operations we performed on
equations in Example 1 that preserve the solution to the system. One such operation was to
multiply an equation, left and right, by a nonzero real number. The other operation was to add
the left side of one equation to the left side of another equation and correspondingly the right
side to right side. These two operations are called row operations because they apply to entire
rows (equations) in the system of equations.
Consider again the system of three equations on which we started the substitution method
before:
A.
X + 4Y + 2Z
= 29,
B.
X + 3Y + 3Z
= 35,
C.
3X + Y + 2Z
= 33.
The goal of the full elimination method is to “eliminate” all coefficients except that of the
X in the first equation, Y in the second equation, and Z in the third equation. That would result
in a system that appears in diagonal form, having coefficients of 1 along the top-left-tobottom-right diagonal and coefficients of 0 everywhere else. The system would then appear
like the following, in which the underlined blank denotes a number to be determined by our
calculations,
Chapter 3
Systems of Linear Equations and Inequalities
56
1X + 0Y + 0Z
= __,
0X + 1Y + 0Z
= __,
0X + 0Y + 1Z
= __.
With the system in diagonal form, we would be able to read the solutions directly as X =
__, and so on. This is precisely how a computer or graphing calculator solves a system of
linear equations using the elimination method.
In the example that follows, we will use the row operations of the elimination method
only as far as is necessary to eliminate the coefficient of X in equations B and C and the
coefficient of Y in equation C. This will result in an equivalent system of equations that looks
like
1X + 4Y + 2Z
= 29,
0X + 1Y + __Z
= ___,
0X + 0Y + 1Z
= ___.
A system of equations like this is said to be in triangular form or row-echelon form
because the coefficients of the system form a diagonal of 1’s from upper left to lower right,
with 0’s in the “triangle” below the diagonal. Notice that, when the system is in triangular
form, the bottom equation says Z = ___. From that point on, we will use the technique of
backward substitution, using the known value of Z in the second equation to solve for Y.
With values for both Z and Y in hand, we solve for X in the first equation.
To summarize, our method will be to use row operations to reduce the system of three
equations to triangular form, having zeros below the main diagonal. Then we will start to read
the solution from bottom to top, substituting known values as necessary. An example will
show how the method works.
Example 2
Elimination Method on Three Equations in Three Variables
Solve the following system by the elimination method:[ED: Tabs removed; not needed for
comp]]
A. X + 4Y + 2Z = 29,
B. X + 3Y + 3Z = 35,
C. 3X + Y + 2Z = 33.
Solution
We begin by eliminating the coefficient of X in equation B. Equation B has a “1” coefficient
on X, so the necessary row operation is to multiply equation A by <->1 and add the result to
equation B: [comp: align each set of eq. on equal signs]
<->1A <->1X – 4Y –2Z = <->29
+B
1X + 3Y + 3Z = 35
= B1
0X – 1Y + 1Z = 6
Likewise, we must eliminate the coefficient of X in equation C. Multiply equation A by <>3 and add that to equation C, getting the result,
<->3A <->3X – 12Y – 6Z
= <->87
+C
3X + 1Y + 2Z
= 33
= C1
0X – 11Y – 4Z
= <->54
Chapter 3
Systems of Linear Equations and Inequalities
57
To get a good look at our progress in the elimination method, we now rewrite the system
of equations using B1 and C1 in place of the original B and C:
A.
1X + 4Y + 2Z = 29,
B1.
0X – 1Y + 1Z = 6,
C1.
0X – 11Y –4Z = <->54.
Clearly, the two row operations we just performed have taken us most of the way toward
the triangular form. The next step is to eliminate the coefficient of Y (<->11) in the third row.
A row operation using equation A applied to equation C1 would create a nonzero coefficient
on the X variable in C1, which we want to avoid. However, row B1 has a 0 coefficient on X, so
multiplying that equation by any number will result in a 0X being added to the 0X in equation
C1, retaining the 0 coefficient that the triangular form calls for. Therefore, we use equation B1
for the row operation of the next step. Multiply B1 by –11, and add it to C1 as follows:
<->11*B1
0X + 11Y –11Z
+ C1
0X – 11Y – 4Z
= C2
0X +0Y <-> 15Z = <->120
= <->66
= <-> 54
To complete our work on C2, we divide both sides of the equation by <->15,
C3.
0X + 0Y + 1Z
= 8.
Now, turn to equation B1. We want the coefficients of X and Y to be 0 and 1, respectively.
Multiplying by –1 will give us an equation B2 with the desired coefficients:
B2.
0X + 1Y – 1Z = <->6.
Collecting the new equations B2 and C3 with equation A, we have the system,
A.
1X + 4Y + 2Z = 29,
B.
0X + 1Y – 1Z = <->6,
C3.
0X + 0Y + 1Z = 8,
X+
or
4Y +
2Z
Y–Z
= <->6;
Z
= 8.
This system is in triangular form. Use back-substitution to solve the system.
Substituting Z = 8 into equation B2, we solve for Y:
Y – 8 = -6,
Y = 2.
Substituting Y = 2 and Z = 8 into equation A, we solve for X:
X + 4(2) + 2(8) = 29,
X + 24 = 29,
X = 5.
CHECK THE SOLUTION
A. 1X + 4Y + 2Z = 29
5 + 4(2) + 2(8) = 29
5 + 8 + 16 = 29 ✓
B. 1X + 3Y + 3Z = 35
5 + 3(2) + 3(8) = 35
5 + 6 + 24 = 35 ✓
C. 3X + 1Y + 2Z = 33
= 29;
Chapter 3
Systems of Linear Equations and Inequalities
58
3(5) + 2 + 2(8) = 33
15 + 2 + 16 = 33 ✓
Make a Note
For solving a system of three or more equations with the same number of
variables, the elimination method is used to transform the system of
equations into triangular form. That reveals the value of one variable, and
then the other variables can be calculated by backward substitution.
Elimination Method in Matrix Form
The elimination method consists of elementary row operations (multiplying an equation by a
nonzero number, adding one equation to another) applied to the coefficients and right<->hand
side of a system of equations. The names given to the variables, such as X, Y, or Z do not
matter. In fact, we will now see that it is easier to work the elimination method when the
system of equations is written as an array of coefficients and right-hand-side values without
the variable names.
A matrix is a rectangular array of numbers. A system of equations can be written in
matrix form by arraying the coefficients and right-hand-side numbers in their natural place.
For example, the system of equations we used in Example 2 would be expressed in matrix
form, with the rows (equations) labeled R1, R2, and R3 and the right-hand-side numbers
separated from the coefficients by vertical lines, as follows:
1 X + 4Y + 2Z = 29
R1
1 4
2 29
1 X + 3Y + 3Z = 35
R2
1 3
3 35
3 X + 1Y + 2 Z = 33
R3
3 1
2 33
Elementary row operations are easy to perform on the system in matrix form. For
example, to indicate that you are doing the row operation “multiply row 1 by –1 and add to
row 2,” you could simply make the notation “–1R1 + R2” and give the resulting system of
equations in matrix form:
–1R1+R2
1
4
2
29
0
−1 1
6
3
1
33
2
The next row operation in the solution of that example was to multiply row 1 by –3 and add
the result to row 3:[comp: for matrices throughout, align each column of numbers within the
matrix.]
<->3R1 + R3
1
4
2
| 29
0
<->1
1
| 6
0
<->11
<->4
| <->54
The third step was to multiply row 2 by –11 and add to row 3:
<->11R2 + R3
1
4
2
| 29
0
<->1
1
| 6
0
0
<->15
| 120
Chapter 3
Systems of Linear Equations and Inequalities
The final step to triangular form was to multiply row 3 by
1
4
2
<->1R2
0
1
<->1
| <->6
−1
R3
15
0
0
1
| 8
59
−1
and row 2 by –1:
15
| 29
Here, remembering that the columns of the matrix are identified with the variables X, Y, Z, and
the right-hand side, we read part of the solution from the bottom row, Z = 8, and recover the
other solutions by substitution.
Make a Note
Row Operations on a System in Matrix Form
Each row in the matrix form corresponds to one equation. The elementary
row operations are the three transformations that do not change the solution
to the system:
1. Multiplying an entire row by a nonzero number.
2. Adding one row to another.
3. Interchanging the order of the rows in the matrix.
When the elimination method is programmed into a computer or a graphing calculator, the
method continues beyond the triangular form, eventually reducing the matrix so that it has 1’s
on the diagonal coefficients (upper-left to lower-right), 0’s for all other coefficients, and
numbers in the right-hand-side values. The matrix form of our example would eventually look
like
1
0 0
5
0
1 0
2.
0
0 1
8
In that reduced row-echelon form (rref), you can read the solution directly. The first three
columns are for X, Y, and Z, so the solution reads X = 5, Y = 2, and Z = 8.
One of the extraordinary abilities of a graphing calculator is to solve a system of
equations by transforming it through row operations to the rref routine.
Learning with Technology
Using a Graphing Calculator to Solve a System of Equations
1. Set up a blank matrix. After calling up the entry screen for the Matrix tool, choose the
Edit option. You must first select a name for the matrix (we’ll call it "A"). The calculator will
also ask you to define the number of rows (equations) and columns (variables + right-side) in
the matrix. A matrix with 3 rows and 4 columns is called a “3-by-4” matrix, often written in
shorthand as “3  4.” A 3  4 array may not show up entirely on one screen. In the screen
shown here, the underline ("_") symbols in the far right column indicate that more of the array
is off the right side of the screen.
{INSERT FIG. UN3.2a}
2. Edit the system matrix. Enter the coefficients and right-hand-side numbers from the
system of equations into the corresponding cells of the matrix. The example here shows
columns 2, 3, and 4. The _ to the left of the first column indicates that some of the matrix is
off the screen to the left.
{INSERT FIG. UN3.2b}
Chapter 3
Systems of Linear Equations and Inequalities
60
3. Invoke the rref routine. Expanded form moved above by 1st mention}. Return to the
main screen, then call up the reduced row echelon form by invoking the calculator’s built-in
routine for that procedure. It is accessed in the TI-83 model by the name rref as a
Matrix/Math function on the Matrix entry screen. as shown here.
{INSERT FIG. UN3.2c}
After the parenthesis of the rref instruction, insert the matrix name A by entering it from
the Matrix/Names screen:
{INSERT FIG. UN3.2d}
Hitting Enter at this point will invoke the calculator’s rref routine to operate on your
matrix A, giving the matrix form of the solution to the system of equations. In the final screen,
shown here, one has to remember how each column is interpreted: the first three columns are
the coefficients of the variables X, Y, and Z, and the rightmost column is the solution column.
Thus, we read the system as saying 1X = 5, 1Y = 2, and 1Z = 8.
{INESRT FIG. UN3.2e}
We are now in a position to solve the problem that was posed in the Spotlight for this
section. Recall that Carol’s task was to determine how many inquiries to Metropolitan
University’s graduate certificate program in management were generated from the
advertisements in each of four different newspapers and magazines.
Example 3
Metropolitan University’s Advertising
When we left Carol in the Spotlight, she was about to show the assistant director of
Admissions how she could find out how effective his various advertisements had been over
the past four months. Mr. Martinez gave Carol the following information about the history of
the advertisements and the inquiries received.
Source:
Fortune
Chicago
Chicago
Magazine
Crain’s
Chicago
Business
Suburban
papers
Price/ad:
$3,500
$3,700
$2,300
$2,400
Total
monthly
cost
Inquiries
received
August
ads
1
0
1
3
$13,000
520
Septembe
r. ads
0
1
1
2
$10,800
445
October
ads
0
1
0
2
$ 8,500
300
Novembe
r ads
1
0
1
2
$10,600
440
Carol had in mind a linear model for the effects of any one magazine’s advertising. This
means that she was assuming that any one source (magazine or newspaper) would always
produce the same number of inquiries per advertisement, from one month to the next. The
linear model also implies that if an ad appeared 3 times in the suburban papers in August, it
should produce 1.5 times the number of inquiries as when only 2 ads were placed, which was
the case in September, October, and November.
Chapter 3
Systems of Linear Equations and Inequalities
61
We shall use the letters r1 to r4 to represent the response rate for each of these four
publications. The response rate is the number of inquiries per ad, per month, for that
publication. Carol’s linear model of the effects of advertising on inquiries can be written in the
following equation:
r1(Fortune ads) + r2(Chicago Magazine ads) + r3(Crain’s ads) + r4(suburban ads) = total
inquiries.
Under Carol’s assumption of a linear model, this equation will be the same for every
month. In the month of August, for example, Metropolitan University took out 1 Fortune ad, 1
Crain’s ad, and 3 suburban ads, and the result was 520 inquiries. Therefore, Carol’s linear
model supposes that the number of inquiries will satisfy the equation,
r11 + r20 + r31 + r43 = 520 (August inquiries).
Similarly, her model supposes that
r10 + r21 + r31 + r42 = 445 (September inquiries),
r10 + r21 + r30 + r42 = 300 (October inquiries),
r11 + r20 + r31 + r42 = 440 (November inquiries).
Think about This
Carol has made a mathematical model of the process by which
advertisements result in inquiries to the Admissions Office. This model is
only as good (accurate) as the assumptions that go into it. Take a moment to
think about how reasonable her assumptions are. Consider the following
questions that challenge the linear model.
1. Can the effect of an advertisement be different from month to month?
Why or why not?
2. Can the effect of more ads in a month in the same magazine be more or
less than would be predicted by multiplying the assumed “response rate per
ad” by the number of ads? Why might the effect be more? Why might it be
less?
3. Can the effect of advertising in one magazine make the advertising in
another magazine, in the same month, more or less effective than otherwise?
What would make it more effective? What might make it less effective?
As you probably can see, each of Carol’s assumptions is open to some challenge. But
each of the assumptions is still reasonable enough for her to proceed with the linear model and
see if the results make sense, especially since Mr. Martinez doesn’t have any alternative right
now.
Given Carol’s model, solve for each of the response rates to determine the effectiveness
of advertising in each publication.
Solution
Putting the coefficients and totals in matrix form, with a top row reminding us of the variables,
we have
r1
r2
r3
r4
A.
1
0
1
3
520
B.
0
1
1
2
445
C.
0
1
0
2
300
D.
1
0
1
2
440
Chapter 3
Systems of Linear Equations and Inequalities
62
The first step toward getting the matrix into triangular form will be to eliminate the 1
under r1 in the bottom row. We do so by multiplying row A by <->1 and adding the result to
row D. This gives us a new matrix,
r1
r2
r3
r4
A.
1
0
1
3
520
B.
0
1
1
2
445
C.
0
1
0
2
300
D1.
0
0
0
<->1
<->80
That was real progress because we can already see the equation <->r4 = <->80 emerge in
row D1. This will tell us that each advertisement in the suburban papers produces 80
responses. Continuing with the elimination method, we must eliminate the 1 under r2 on row
C. Doing so by the appropriate row operation (<->1  row B added to row C) gives us a new
table:
r1
r2
r3
r4
A.
1
0
1
3
520
B.
0
1
1
2
445
C1.
0
0
<->1
0
<->145
D1.
0
0
0
<->1
<->80
After multiplying rows C1 and D1 by <->1, the matrix will be in triangular form, so we
can start the process of backward substitution to get the four response rates.
Row D1 says <->1  r4 = <->80, so r4 = 80 responses per ad in the suburban papers.
Likewise, row C1 reveals that r3 = 145 responses per ad in Crain’s Chicago Business. Using
row B and our known values of r3 and r4 we determine the value of r2 as follows
B. r2 + 145 + 2(80) = 445,
so
r2 = 140 (responses per ad in Chicago Magazine).
Finally, we use row A to calculate r1,
r1 + 0r2 + 1r3 + 3r4 = 520,
r1 + 0 + 145 + 3(80) = 520,
so
r1 = 135 (responses per ad in Fortune’s Chicago edition).
Learning with Technology
1. To use a graphing calculator to solve this system, what matrix dimensions
would you enter?
2. Verify our solution using your graphing calculator.
Answer: 1. 4  5
The conclusion of this linear model is that the number of responses per ad in Fortune,
Chicago Magazine, Crain’s, and the suburban papers are 135, 140, 145, and 80, respectively.
The first three response rates are quite similar, but the number of responses per ad in the
suburban papers is much less than the others.
This analysis has a very practical value. The original table of values given by Mr.
Martinez shows that Fortune and Chicago Magazine ads cost about the same, but the Crain’s
ad costs much less. When ads in those three magazines produce similar effects in terms of
Chapter 3
Systems of Linear Equations and Inequalities
63
response rate, what does that tell us about where advertising is most effective per dollar
spent? And look at the suburban papers. How does their effect per dollar spent compare with
the other sources?
INCONSISTENT AND DEPENDENT SYSTEMS
Before leaving the topic of systems of linear equations, let's look at two other possible
outcomes of the elimination method. When we solve a system that has the same number of
equations as the number of unknowns, we expect to get a solution where there is one value for
each variable. But in section 3.1 of this chapter we found that there are two situations where
this kind of solution will not occur. As you will recall, if we visualize a system with two
variables and two equations as two lines on the coordinate plane, the system has no solution if
the two lines are parallel and therefore do not intersect. In this case, the equations are
inconsistent with each other. Another possibility is that the two lines will actually be one and
the same, in which case all points on that line are solutions to the system of “two” equations.
That is the situation of a dependent system of equations.
With more than two variables, it is not as easy to visualize the system of equations. A
linear equation in three variables can be pictured as a plane in three-dimensional space. Three
such equations can be drawn as three planes, and if they are all tilted in different directions
they will intersect in a single point in three-dimensional space, which is the solution to the
system of three equations in three variables. But there are various situations where there is no
solution, or no common point of intersection.
Can the elimination method reveal when a system of equations has no solution or has
many solutions? The answer is “yes” because the effect of the row operations that would
otherwise produce a triangular array of coefficients in a table will produce a different result
instead. Let’s look at two examples.
Example 4
Inconsistent System
An inconsistent system has no solution. As an example, attempt to solve the following system
of equations using the elimination method:
A. 2X + 3Y + Z = 25,
B. 4X + Y + 5Z = 40,
C. 2X – 2Y + 4Z = 10.
Solution
To work the elimination method most rapidly, we put this system in matrix form, where the
columns correspond to the variables X, Y, and Z and the right-hand sides of the equations:
A.
B.
C.
2
4
2
3
1
<->2
1
5
4
25
40
10
In the first sequence of steps, we will multiply row A by 1/2 so that X has a coefficient of
1 on the diagonal. Then we will multiply the new first row A1 by <->4 and add it to row B.
Next, we multiply A1 by <->2 to add to row C to eliminate the X coefficients in those two
rows. The result of those three operations is the following:
A1.
B1.
C1.
1
0
0
1.5
<->5
<->5
0.5
3
3
12.5
<->10
<->15
In the matrix above, we can see trouble emerging. Rows B1 and C1 have identical
coefficients but different right-hand-side values. It will be impossible to find values of Y and Z
Chapter 3
Systems of Linear Equations and Inequalities
64
that satisfy both equations simultaneously. As we proceed with the elimination method, this
will become even more apparent. Continuing, we divide row B1 by <->5 to produce a 1
coefficient for Y on the diagonal B2. Then we multiply B2 by 5 and add the result to row C1 to
get the following table:
A1.
B2.
C2.
1
0
0
1.5
1
0
0.5
12.5
<->0.6 2
0
<->5
The last equation says that 0X + 0Y + 0Z = <->5, which is impossible for any real
numbers X, Y, and Z to satisfy.
Make a Note
The appearance of a row of zeros in a table of coefficients, together with a
nonzero right-hand-side value, is the signal that the system of equations has
no solution.
Example 5
Dependent System
A dependent system has more than one solution. This is indicated when two rows become
identical, which means they have the same coefficients and the same right-hand-side value. At
the end of the elimination method, this results in a row of all zeros. Consider this example:
A.
2X + 3Y + Z
= 25
B.
4X + Y + 5Z
= 40
C.
2X – 2Y + 4Z
= 15
We put this system in matrix form.
A.
B.
C.
X
2
4
2
Y
3
1
<->2
Z
1
5
4
25
40
15
Multiply row A by 1/2, then add –4 times the new row A1 to row B, and <->2  A1 to row
C in order to eliminate the X coefficients on rows B and C The result of those three operations
follows:
A1.
B1.
C1.
X
1
0
0
Y
1.5
<->5
<->5
Z
0.5
3
3
12.5
<->10
<->10
Divide row B1 by <->5 to produce a Y coefficient of 1 on the diagonal B2. Then multiply
that row by +5, and add the result to row C1 in an attempt to eliminate the coefficient of Y
(C2). We get the following matrix:
A1.
B2.
C2.
X
1
0
0
Y
1.5
1
0
Z
0.5
<->0.6
0
12.5
2
0
The last equation says that 0X + 0Y + 0Z = 0, which can be satisfied by many different
values of X, Y, and Z.
Chapter 3
Systems of Linear Equations and Inequalities
65
Make a Note
The appearance of a row of all zeros in the table, including a zero for the
right-hand-side value, is the signal that the system of equations has more
than one solution.
In this chapter, we will not consider how to treat these two special cases any further.
However, in a course on matrix algebra (linear algebra) you would learn more about these,
including how to “solve” a system of equations that has no solution and how to describe all the
solutions to a system that has many solutions.
SUMMARY
This section of the text introduced a topic that has extremely broad usefulness across many
fields of study and professions. The application exercises in this section will demonstrate the
range of uses for systems of linear equations.
The topic is easy to describe: you have several different relationships among the same set
of variables. Each relationship corresponds to an equation, so the mathematical description of
the situation is a system of several equations using several variables.
In this section, we looked only at the problem where the number of variables is the same
as the number of equations and all the equations are linear. This problem almost always has
exactly one solution, which you can calculate using the elimination method.
Each step of the elimination method is an elementary row operation, which can be the
multiplication of a row by a nonzero number, the addition of one row to another, or an
interchange of the order of rows in the matrix. Elementary row operations do not change the
solution to the system, yet they simplify the system of equations. The end result of the
elimination method is a system of equations in reduced row-echelon form, where the only
nonzero coefficients are 1’s along the main diagonal of the matrix (from the upper-left, down
and to the right). In that form, it is possible to read the solution of the system directly from the
right-most column of the matrix.
EXERCISES
Skill Builders
Problems 1–5 should be solved in sequence using the following system of equations.
A. <->X
B. 3X
C. 2X
+ Y + 2Z
=3
– 2Y
+ 4Z = 19
+ 3Y
– 5Z = <->19
1. What operation on row A would enable you to eliminate the coefficient of X in row B?
Perform that operation to obtain a new row B, and label it B2.
2. What operation on row A would enable you to eliminate the coefficient of X in row C?
Perform that operation to obtain a new row C, and label it C2.
3. What operation on row B2 would enable you to eliminate the coefficient of Y in row C2?
Perform that operation to obtain a new bottom row, and label it C3.
4. What value of Z is implied by equation C3? Substitute that value of Z in equation B2, and
solve for Y.
5. Substitute the values of Z and Y found in problem 4 into equation A and solve for the
remaining unknown, X.
Problems 6 – 10 should be solved in sequence using the following system of equations.
Chapter 3
Systems of Linear Equations and Inequalities
66
A. X + Y
B. <->X
C. 2X
+ 2Z
+2Y
+ 3Y
=2
+ 4Z = <->2
– 4Z = <->1
6. What operation on row A would enable you to eliminate the coefficient of X in row B?
Perform that operation to obtain a new row B, and label it B2.
7. What operation on row A would enable you to eliminate the coefficient of X in row C?
Perform that operation to obtain a new row C, and label it C2.
8. What operation on row B2 would enable you to eliminate the coefficient of Y in row C2?
Perform that operation to obtain a new bottom row, and label it C3.
9. What value of Z is implied by equation C3? Substitute that value of Z in equation B2 and
solve for Y.
10. Substitute the values of Z and Y found in problem 4 into equation A, and solve for the
remaining unknown, X.
Use the elimination method to solve the following systems of equations. Check your answer by
substituting values back into the original equations.
11. A. 1X + 0Y + 1Z = 5
B. 0X + 1Y – 2Z = <->10
C. 2X + 5Y + 3Z= 4
12. A. 1A + 2B + 1C = 7
B. –1A
– 2B + 4C
C. 0A + 2B + 3C= 8
=3
13. A. 1A + 2B – 1C = 7
B. –1A
+ 5B + 2C = 15
C. 2A – 6B + 1C= <->13
14. A. <->3X + 3Y – 5Z = <->6
B. 2X – 1Y + 2Z = <->1
C. –1X
+ 0Y + 3Z = 13
15. A.
B.
C.
16. A.
B.
C.
1A + 3B + 1C = 7
0A + 5B + 2C = 11
2A + 4B – 3C= <->3
1X + 0Y + 1Z = 3
1X – 1Y + 1Z= 0
3X – 2Y + 5Z= 11
17. A. 2A + 3B + 4C = 0
B. 5A + 6B + 8C = 0
C. 4A – 2B + 3C= 0
18. A. 1A + 1B + 1C= 3
B. 2A + 3B – 1C= – 2
C. 3A + 0B + 1C= 0
19. A. 1X+ 2Y + 1Z = 7
B. 1X – 1Y + 1Z = – 2
C. –1X
+ 1Y – 2Z
=6
20. A. 1A + 1B – 1C= 0
B. 0A + 1B + 1C = 0
C. 1A + 2B + 2C= 6
For each of the following determine whether the system is inconsistent (has no solution) or
dependent (has more than one solution).
Chapter 3
Systems of Linear Equations and Inequalities
67
2 X + 3Y + 4Z = 1
21. 4 X + 6Y + 8Z = −1
4 X − 2Y + 3Z = −3
1X + 2Y − 4Z = 11
22.
− 2 X + 3Y − 6Z = 6
3 X + 4Y − 8Z = 13
1X + 1Y − 1Z = 0
23. 0 X + 1Y + 1Z = 0
1X + 2Y + 0Z = 0
Applications and Analysis
When you solve a system of equations in the following section, use the elimination method.
Show clearly which equation or table row you are multiplying by what number and adding to
which other equation as you proceed through the steps of the elimination method.
24. Metropolitan University (part B). Pretend that it is the first week of December.
Carol Haines is in the Admissions Office going through the mail, and Mr. Martinez walks in.
“Carol, am I glad to see you. Say, I noticed on my November financial report that we had
spent $13,000 on advertising rather than the $10,600 that I told you earlier in the week. I
checked, and it turned out that we had run 3 ads in the suburban papers, not 2. And on top of
that, my secretary just brought in a batch of inquiries that had come in the mail while she was
ill last week. Her temporary replacement didn’t know what to do with them and just set them
aside. There are 40 inquiries in that batch, which brings the November total to 480, not 440.
How much of a difference do you think these changes will make in your estimates of the
effectiveness of advertising in each of our sources?”
“That depends,” she replies. “I didn’t run a sensitivity analysis on the original model. That
is, I never tried to change the numbers slightly to see what effect such changes would have.
But the problem was rather quick to solve. It will take me only a minute to change those two
numbers and give you the results. Do you want to see what it looks like?”
“Please,” said Mr. Martinez.
Carol changed the November inquiries from 440 to 480 and the number of suburban ads
in November from 3 to 2. This gave her the table shown here.
r1
r2
r3
r4
1
0
1
3
520
0
1
1
2
445
0
1
0
2
300
1
0
1
3
480
Try to solve this system of equations using the elimination method. What has happened?
25. Restaurant staffing. Many restaurants are open long hours during the day, but only in
certain times of the day is the restaurant extremely busy. As a consequence, a restaurant owner
will typically hire waiters on several shifts, and the shifts are designed to overlap during the
busy periods. In that way, the restaurant has many waiters on duty during the busy periods but
fewer waiters on duty during the slow periods.
Chapter 3
Systems of Linear Equations and Inequalities
68
Suppose that Fred’s Seafood Restaurant is open from 11 a.m. to 11 p.m., serving lunch
and dinner. Fred Trawler, owner of the restaurant, has set up four shifts for his waiters. In the
table here, those shifts are denoted by a 1 in any time period that the shift’s staff are supposed
to be on duty. A 0 indicates that the shift members are not on duty during that time period. A
zero separating two 1’s down a column indicates a “split shift,” in which the waiter is given
time off but is asked to come back to the restaurant later in the day. Fred has made the split
shifts shorter in total hours than regular shifts due to the inconvenience of having to go off
duty and come back on duty. Shift 4 is for part-time waiters, who are usually students at a
nearby university.
The rightmost column of the table also shows how many waiters will be required during
each three-hour block of time during the work day.
Time block
Waiters
Shift 1 Shift 2 Shift 3 Shift 4
required
11-2 p.m.
1
1
0
0
9
2–5 p.m.
1
0
1
0
4
5–8 p.m.
1
1
1
1
14
8–11 p.m.
0
0
1
0
3
A. Write an equation that says, “The total number of waiters on duty between 2 and 5 p.m. is
equal to 4.” (Hint: Write S1, S2, etc., for the number of waiters assigned to shift 1, shift 2, etc.)
Which shifts are on duty during 2-5 p.m.? Use the variable names S1, S2, etc., to write an
expression for the total number of waiters on duty during 2-5 p.m. [nice edits}
B. Use the elimination method to find the numbers of waiters that Fred should hire on each of
the four shifts. If he must have whole numbers of waiters (not fractions), how many should he
hire?
C. A new enterprise has located nearby, and Fred has noticed a much larger lunch crowd
together with a slightly larger evening crowd. He has decided he now needs 12 waiters for the
11-2 time block, 4 for the 2-5 time block, 15 for the 5-8 time block, and 3 for the 8-11 time
block. Use the elimination method to find the number of waiters that Fred should hire for each
shift.
26. Advertising effects and walk-in traffic. Lois Waxman is the owner of Somebody
Cares Gift Shop. She advertises on KABC radio and in the local newspaper. But she also gets
a lot of walk-in traffic because her store has been around for a while and the large sign out
front reminds people where she is. However, she is not sure how much of her sales are due to
walk-ins and how much is due to the advertising she is paying for. Over 3 months, Lois
collected the following information that might help her understand where her sales are coming
from.
January: 2 ads on KABC, 3 in the local paper, and sales of $10,000
February: 1 ad on KABC, none in the local paper, and sales of $6,000
March: 1 ad on KABC, 2 in the local paper, and sales of $7,500.
A. Let s1 denote the dollar sales per ad in KABC, s2 the dollar sales per ad from the local
paper, and s3 the unknown amount of sales from foot traffic. Explain why the equation, 2s1 +
3s2 + 1s3 = 10,000 could be used to account for the sales in January.
B. Write equations to account for the sales in February and in March.
C. Solve the system of three equations in three unknowns. How much of Lois’s sales are due
to walk-in traffic?
27. Emergency room medical staff. A hospital’s emergency room is open 24 hours a
day. At Lakeview Hospital, the medical staff work 12-hour shifts 3 days a week. The shifts are
6 a.m. to 6 p.m., noon to midnight, 6 p.m. to 6 a.m., and midnight to noon. Experience has
shown that the need for emergency medical staff is greatest during the noon to 6 p.m. period,
Chapter 3
Systems of Linear Equations and Inequalities
69
due to work-related accidents and traffic emergencies. In the table, a 1 in the shift columns
indicates that the staff on that shift are working during a particular period. The rightmost
column shows the need for staff during the period.
Staff
Shift 1 Shift 2 Shift 3 Shift 4
required
6 a.m.–noon
1
0
0
1
21
noon–6 p.m.
1
1
0
0
35
6 p.m.–midnight
0
1
1
0
32
midnight–6 a.m.
0
0
1
1
18
A. Write an equation that says, “The total number of medical staff on duty between 6 a.m. and
noon is equal to 21.” (Hint: Write S1, S2, etc. for the number of staff assigned to shift 1, shift 2,
etc. Which shifts are on duty during the period 6 a.m. to noon?)
B. Use the elimination method to try to find the numbers of medical staff that Lakeview
Hospital should hire on each of the four shifts. What has happened?
C. One way to eliminate the dependency in this problem is to modify the coefficient matrix so
that not all shifts work continuous 12-hour blocks. But we shouldn’t split shifts that are this
long (6 hours on, 6 off, 6 on, 6 off would be a killer.) So shorten one of the shifts to a single 6hour segment. What then is the solution to the assignment problem? Can you explain its
peculiar feature?
Analysis
Interchanging rows and columns in the elimination method. Look back to the
Metropolitan University example. We gave letter names to the variables describing responses
per advertisement in the Fortune (r1), Chicago Magazine (r2), Crain’s (r3), and suburban
newspapers (r4). With that notation, the equations in the system looked like this:
August: 1r1 + 0r2 + 1r3 + 3r4 = 520,
September: 0r1 + 1r2 + 1r3 + 2r4 = 445,
October: 0r1 + 1r2 + 0r3 + 2r4 = 300,
November: 1r1 + 0r2 + 1r3 + 2r4 = 440.
28. If we interchange the rows of this system of equations, it will change the appearance
of the table of numbers that we start with to work the elimination method. For example, the
system might be rearranged like this.
September: 0r1 + 1r2 + 1r3 + 2r4 = 445,
August: 1r1 + 0r2 + 1r3 + 3r4 = 520,
November: 1r1 + 0r2 + 1r3 + 2r4 = 440,
October: 0r1 + 1r2 + 0r3 + 2r4 = 300.
Does rearranging the rows change the solution? If so, why? If not, why not? (Hint: To answer
this question, don’t try to solve this system. Think about what it means to rearrange the rows
and about the meaning of a solution to the system of equations.)
29. We could also have chosen a different order to write the variables in. For example,
instead of the order Fortune, Chicago, Crain’s, suburban, we might have written it in the
order Crain’s, suburban, Fortune, Chicago. In this order, the equations would appear as
follows:
September:
1r3 + 2r4 + 0r1 + 1r2 = 445,
August: 1r3 + 3r4 + 1r1 + 0r2 = 520,
November: 1r3 + 2r4 + 1r1 + 0r2 = 440,
Chapter 3
Systems of Linear Equations and Inequalities
70
October: 0r3 + 2r4 + 0r1 + 1r2 = 300.
Rearranging the order of the variables in a system of equations has same effect as
interchanging the sequence of columns of the table that we use for the elimination method.
Does rearranging the columns of the table in this way change the solution? If so, why? If not,
why not? (Hint: Again, don’t try to solve this system. Think about what it means to rearrange
the variables, and what is the meaning of a solution to the system of equations.)
30. Rearranging rows and columns can put this table into upper-triangular form simply
and without any calculations. In this example, there may be numbers other than 1 along the
diagonal, but the table will have a triangular form.
A
B
C
D
W
0
1
0
6
X
2
4
7
3
Y
5
0
0
<->2
Z
0
2
0
0
100
180
140
90
A. Find a way to interchange rows and columns so that a triangular form appears. (This is like
working a puzzle. Hint: how many zeros should there be at the bottom row of the table in
triangular form, and in which columns should they be placed? What about the next-to-bottom
row?)
B. When you have the table in triangular form, complete the solution of the system by
backward substitution.
Not all systems of linear equations can be prepared for quick solution in this way, but if
you see a lot of zeros in a system table, it is worth taking a minute to try a few row or column
switches.
Graphing Calculator Exploration
31. Real estate prices. Daniel Marlin is a real estate agent in the seashore community of
Bayside. His observation of the real estate market suggested to him that three features
determine property prices: the floor area of the house, the amount of shoreline, and the
acreage of the property. Daniel collected the following data on three houses that sold recently
in Bayside.
House
Watson
Carlisle
Becker
Floor
(sq. ft.)
1,000
800
1,500
Shoreline
(ft.)
75
50
100
Land
(acre)
1
0.75
3
Selling
price ($)
110,000
85,250
185,000
Use a graphing calculator to solve this system in matrix form. Write a linear equation that
can be used to predict the selling price of any house in Bayside.
Spreadsheet Exercise
32. The Microsoft ExcelTM spreadsheet program can solve systems of simultaneous linear
equations. Excel is installed on so many computers and is used for so many purposes in
business that it is a good program to know.
Excel has a Solver routine to solve systems of linear equations as well as many other
types of mathematical problem. The Solver is found under the Tools menu.  Solve Exercise
18 above using the Solver, as follows.
A
B
C
D
E
F
1
0
0
0
Chapter 3
2 House
3 Watson
4 Carlisle
5 Becker
Systems of Linear Equations and Inequalities
71
Floor (area)
Shoreline Land
Selling
(ft.)
(acre)
price ($)
1,000
75
1 110,000
800
50
0.75
85,250
1,500
100
3 185,000
A. Set up the system of linear equations in matrix form beginning in row 2. Then add a row of
zeros above Floor, Shoreline, and Land in cells B1 to D1, as shown. These cells will act as the
unknown “prices” for Floor, Shoreline, and Land.
B. Create a formula for cell F3 that would equal the price of the Watson property if the prices
of floor, shoreline, and land are the values in cells B1 to D1. That is, for cell F3, enter
= B$1*B3+C$1*C3+D$1*D3.
The dollar signs in B$1 to D$1 will instruct Excel to hold row 1 when this formula is copied
to other cells in the next step. A zero will appear in cell F3 because the current values of B1 to
D1 are all zero.
C. Copy the formula from cell F3 to cells F4 and F5. Zeros should appear in those cells, too.
D.
Call up the Solver from the Tools menu. The Solver Parameters box will appear.
Clear the “Set Target Cell” and “Value of:” boxes. In the “By Changing Cells”
box, type B1:D1. To the right of the “Subject to the Constraints” box is an Add button. Click
on this. An “Add Constraint” box will appear. For “cell reference” give F3. Choose = from
the menu. And for “constraint” give E3. This tells the solver to set the price of the Watson
house (F3) equal to the number 110,000 that is in cell E3.
Click “OK,” and the Solver parameters box will appear with one constraint in it,
expressed as $F$3 = $E$3.
Click the Add button
again and use the same procedure to define the constraint F4 = E4. Click “OK.” Click
the
Add button a third time and define the constraint F5 = E5. Click “OK.”
E. Now, in the upper right of the Solver Parameters box, click “Solve.” The solutions to this
system of linear equations will appear in place of the zeros in cells B2 to D2.
Save the spreadsheet in its present form. You can use it to solve any other 3  3 linear
system simply by changing the numbers and labels in the table.
CASE STUDY — APPRAISING REAL ESTATE
A property owner who wants to sell a house often asks a real estate professional to determine
a fair market price for, or appraise, the property. The usual method of a real estate appraiser is
to find several houses that have sold recently in the area and that are comparable to the subject
property on such features as the number of bedrooms and bathrooms, size of the house,
existence of a garage, size of the lot, and so on. By looking at the selling prices of those
comparables, the appraiser can come pretty close to estimating a fair market value for the
subject property.
{INSERT Photo 3.4}
However, it is difficult, if not impossible, for the appraiser to find several houses that are
exactly like the subject property and that have sold recently in the area. The appraiser must
therefore make adjustments for slight differences between each comparable and the subject
property. A comparable may have one more bedroom and a garage, another comparable may
have one less bathroom, and so on. The appraiser has some sense of how much an extra
bedroom, bathroom, or garage should be worth, and they use their sense of the “prices” for
those features to make the necessary adjustment.
As it is commonly practiced, real estate appraisal is rather judgmental and subjective.
However, a mathematical analysis of sold properties can reveal the “prices” of the various
features that the real estate appraiser is trying to take into consideration. In this case study,
we’ll look at data on houses in Greenfield, Ohio, that were about to be sold and use that to
Chapter 3
Systems of Linear Equations and Inequalities
72
estimate a selling price for one particular subject property. The data for this case study were
taken from the web site of the Multiple Listing Service of Greater Cincinnati
<http://www.cincymls.com/>.
The subject property will be 11892 Maple Trail in Greenfield, Ohio. It has 3 bedrooms, 3
other rooms, 1 bathroom, a 1-car garage, and sits on 0.4 acres of land. Five comparables, on
which other buyers had made offers (presumably at the asking price) are listed in the table.
Other
Garage Land
Location
Bedroom rooms Baths (cars) (acres) Price
s
($)
12386 Cameron Dr.
2
2
1
1
0.2
56,000
330 4th St.
2
3
1.5
0
0.6
56,000
545 McKell Ave.
2
3
1.5
1
0.3
62,900
11395 Rowe Rd.
3
5
2
0
0.7
79,900
836 Jefferson St.
3
5
1
2
0.2
95,900
A. Your first task is to create a linear equation model for the price of a property in relation to
the features of the property. Problem 14 of section 3.2 and problem 18 in this section use the
same principle. The price of a property is written as a linear equation with five variables,
which are the five features of the house given in the table above. The coefficient of each
variable in the equation can be interpreted as the “price” of that feature. Using the data in the
table, you can set up a system of five equations with the five unknown “prices.” Solving that
system, you find the price for each feature, which gives you an equation for the price of any
property.
B. Do the “prices” you get for each feature seem reasonable? In particular, what is the price
per acre for land in Greenfield, Ohio? What is the “price” for a garage (per car)? How much
extra would a house cost if it had an extra bathroom?
C. Use the equation from part A to predict the selling price of the subject property at 11892
Maple Trail. The owners listed it at (were willing to sell it for) $65,000. Do you think that was
too much, too little, or just about right?
D. The predictive accuracy of your mathematical model of the property price is limited by the
range of features that it considers. What other features of a property might affect the selling
price?
Chapter Review
The knowledge in this chapter is very practical. We began with the search for two unknowns
that are related by two equations. The range of applications that can be written as a system of
two equations in two unknowns includes demand and supply in economics, production
constraints in management, and problems of parameter estimation and “curve fitting” in
business, biology, psychology, and medicine.
We used the word, "system" to describe the equations that we were solving. This word is used
widely in mathematics. In Chapter 1, we encountered the “real number system.” In this chapter
we have been studying “systems of equations.” The common meaning of the word "system" in
all these instances is that of a collection or set that has some relationship among the parts. This
meaning is important because when a problem can be written as a system of equations, the
solution will be the values of several variables that solve each equation simultaneously.
Graphically, each linear equation in a two-variable system corresponds to a line in the
coordinate plane. So a pair of numbers that solves both equations correspond to a point in the
coordinate plane that lies at the intersection of the two lines. Therefore, the graphical method
to solve a 2  2 system of equations is to plot the lines on a coordinate plane and determine the
coordinates of the point of their intersection. It is often faster, however, to solve the system
algebraically. For 2  2 systems, we used the substitution method to find a solution in three
steps.
Chapter 3
Systems of Linear Equations and Inequalities
73
The solution to an inequality is quite different from the solution to an equation.
Graphically, the solution of a linear equation is a line in the coordinate plane. The graphical
solution to an inequality is a half of the coordinate plane that is bounded by (and may or may
not include) the corresponding line.
Another difference between equations and inequalities is that, whereas a system of
equations in two variables can be solved only if there are exactly two equations, a system of
inequalities can have a solution even if there are many inequalities to be satisfied. We noted in
section 3.3 that a solution set may be a bounded region of the plane or an unbounded region,
or there might be no solution at all. And we observed that some constraints in a system can be
redundant, meaning that they can be added or removed without altering the solution. We also
discussed the concept of slack in a constraint at a point and learned that slack tells us how
much the right-hand side of a constraint can be changed before the constraint becomes
binding.
In the final section of this chapter, we used the elimination method to solve systems of
linear equations that have more than two variables. To solve systems most efficiently, we
performed elementary row operations on matrices consisting of the constants in the equations.
A linear system that has the same number of equations as the number of variables ordinarily
has exactly one solution. However, we did observe that an inconsistent system has no solution,
revealed when two rows with identical coefficients have different solutions. In addition, we
noted the possibility of infinite solutions in a dependent system, where row operations result
in a row of zeros.
CUMULATIVE TEST FOR CHAPTERS 1-3
1. Stockbroker earnings. A stockbroker in New York City is paid a base wage of $2,000
per month plus 30% of the commissions paid by her clients. The commission is 2% of the
dollar amount of the clients’ stocks sold.
A. Write an equation for the stockbroker’s total monthly earnings in relation to the dollar
amount of stocks sold. Express the equation in simplest terms.
B. Graph the equation for the dollar amount of clients’ stock sold ranging from zero to
$500,000.
C. Determine algebraically how much volume (dollar amount of sales) a stockbroker would
have to generate in a month in order to earn $3,500 in total compensation? Verify your answer
by showing that point on your graph.
2. Stockbroker earnings. Another stockbroker made the comment over lunch that his
firm pays him 40% of his commissions and charges clients 2% of the dollar amount of stocks
sold. However, he says he has to hustle each month because his base salary is quite low. Last
month, he says, he earned $4,500 and sold $375,000 worth of stock for his clients. What is his
base salary?
3. Monthly mortgage. The monthly payment on a mortgage loan for a house depends on
the interest rate of the loan and the price of the house. For a $100,000 house loan, the monthly
payment at an interest rate of 5% per year would be $536.82. At a 6% annual interest rate, the
payment would be $599.55 per month.
A. Create a linear equation that shows the monthly mortgage payment in relation to the
interest rate.
B. Using the linear equation from part A, estimate the monthly payment that would be
required if the interest rate were 5.625% per year.
4. Temperature. The equation for temperature in degrees Fahrenheit (F) in relation to
degrees centigrade (C) is F = 32 + 1.8C. What is the equation for the temperature in degrees
centigrade, given the Fahrenheit temperature?
Chapter 3
Systems of Linear Equations and Inequalities
74
5. Materials budget. The cost of the materials used in manufacturing one table is $120
and for one bookcase it is $45.
A. Write an equation that shows the number T of tables and number B of bookcases that can be
manufactured within a materials budget of $8,100, then graph that equation.
B. Determine algebraically how many tables could be manufactured within a $8,100 materials
budget if the firm had already committed to manufacturing 100 bookcases, then show your
answer on the graph from part A.
6. Break-even analysis. A firm has to pay $6,500 per month on utilities and rent, even if
it is not producing anything. The cost of materials, energy and labor used to create its product
is $30 per unit produced.
A. Write an equation that shows the total expenditure E of the firm in relation to the quantity
Q of the product it manufactures, then graph that equation for values of Q between 0 and 200.
B. The firm obtains sales revenue by selling the product for $85 per unit. Draw the firm’s
revenue line on your graph from part A. Determine both algebraically and graphically what
quantity the firm must produce per month in order to break even.
7. Diet planning. Kalista is contemplating a diet composed entirely of raisins and green
peas. The table describes the amount of several ingredients in these two foods and the
recommended daily allowances.
Content per
Raisins (1
Green peas (1
Recommended
piece
cup, or 145 g) cup, 160 grams) daily allowance
Calories
435
125
Maximum 2,000
Protein (grams)
5
8
Minimum 46 g.
Calcium
71
38
Minimum 1,200
mg.
Iron (mg.)
3.0
2.5
Minimum 15
mg.
Vitamin C
5
16
Minimum 60
mg.
If Kalista wants to eat a diet that gives her exactly the minimum requirements of iron and
Vitamin C, how many cups of raisins and how many cups of peas should she eat?
8. Admissions screening. The policy of Valley College is to admit an applicant
automatically if their average SAT score is at least 550 and their high school GPA is at least
2.75, and 100 times the GPA added to the average SAT score is at least 925.
A. Write the conditions for automatic acceptance in the form of a system of linear inequalities.
B. Sketch the “automatic acceptance” region in the SAT-GPA plane.
C. Are any of the constraints redundant?
D. Is the automatic acceptance region bounded or unbounded?
E. If an applicant has a GPA of 3.05 and an SAT of 600, how much slack will they have in
each constraint? Will they be automatically admitted?
9. Advertising effectiveness. Randy and Martin have obtained a franchise to sell
AmritaTM aroma diffusers in the St. Louis region. In their first 3 months of marketing, they
have been using several media. Some people have commented that they have seen the Amrita
ads in a local newspaper. Radio advertising also seems to be somewhat effective.
Advertisements in a citywide magazine also have been producing some results. Randy and
Martin take orders over a toll-free number that they publicize in their ads, but they have not
been asking callers how they heard about the aroma diffuser. Thus, it seems difficult to know
which advertisements are more effective. This table shows the number of advertisements
Chapter 3
Systems of Linear Equations and Inequalities
75
taken per month in each medium together with the monthly total of calls to the toll-free
number. Help Randy and Martin determine the effectiveness of their advertising.
Advertisements
Radio
Newspaper
Magazine
Total calls
February
1
3
2
500
March
0
2
2
380
April
2
4
1
500