SECTION
2.4
Systems of Linear
Equations
Copyright © Cengage Learning. All rights reserved.
Learning Objectives
1 Determine the solution to a system of equations
algebraically, graphically, and using technology
and interpret the real-world meaning of the
results
2 Use the substitution and elimination methods to
solve linear systems that model real-world
scenarios
3 Determine if systems of linear equations are
dependent or inconsistent and explain the
real-world meaning of the results
2
Systems of Linear Equations
3
Systems of Linear Equations
A system of equations is a group of two or more
equations. To solve a system of equations means to find
values for the variables that satisfy all of the equations in
the system.
Systems of equations can involve any number of equations
and variables; however, we will limit ourselves to situations
containing two variables in this section.
4
Systems of Linear Equations
A solution of a pair of linear equations is an ordered
pair of numbers that satisfies both equations. The
ordered pair (5, 3) is a solution of the linear
equations below.
© 2010 Pearson Education, Inc. All rights reserved.
Section 8.1, Slide 5
Systems of Linear Equations
Although (7, 2) makes the first equation true in the
system…
(2)(7) + (4)(2) = 22
…it does not make the second equation true.
7 – (6)(2)  5
Therefore, it is not a solution for the system.
© 2010 Pearson Education, Inc. All rights reserved.
6 6
Section 8.1, Slide
Solving Systems Using the
Elimination Method
• Example: Solve the system.
© 2010 Pearson Education, Inc. All rights reserved.
(continued on next slide)
7 7
Section 8.1, Slide
Solving Systems Using the
Elimination Method
• Example: Solve the system.
• Solution:
Multiply the top equation by 3 and the bottom
equation by 2 to get opposite coefficients for y.
© 2010 Pearson Education, Inc. All rights reserved.
(continued on next slide)
8 8
Section 8.1, Slide
Solving Systems Using the
Elimination Method
Next add corresponding sides of both equations and
the y drops out.
Solve for x.
© 2010 Pearson Education, Inc. All rights reserved.
(continued on next slide)
9 9
Section 8.1, Slide
Solving Systems Using the
Elimination Method
To find y, substitute 2 for x in either equation of the
original system.
Thus, the solution for this system is (2, 1) (Case 1).
© 2010 Pearson Education, Inc. All rights reserved.
1010
Section 8.1, Slide
X -2
X3
11
Solving Systems Using the
Elimination Method
• Example: Solve the system
© 2010 Pearson Education, Inc. All rights reserved.
(continued on next slide)
1212
Section 8.1, Slide
Solving Systems Using the
Elimination Method
• Example: Solve the system
• Solution:
Multiply both sides of the top equation by 4 to clear
the fractions.
© 2010 Pearson Education, Inc. All rights reserved.
(continued on next slide)
1313
Section 8.1, Slide
Solving Systems Using the
Elimination Method
Multiply the bottom equation by 2 and add the
equations to eliminate x from the system:
There are no points common to both lines (Case 2).
A system that has no solutions is said to be
inconsistent.
© 2010 Pearson Education, Inc. All rights reserved.
1414
Section 8.1, Slide
Solving Systems Using the
Elimination Method
• Example: Solve the system
© 2010 Pearson Education, Inc. All rights reserved.
(continued on next slide)
1515
Section 8.1, Slide
Solving Systems Using the
Elimination Method
• Example: Solve the system
•Solution:
Multiply the top equation by 10 and the bottom
equation by 100 to get rid of the decimals.
© 2010 Pearson Education, Inc. All rights reserved.
(continued on next slide)
1616
Section 8.1, Slide
Solving Systems Using the
Elimination Method
Multiply the top equation by –5, and add the two
equations to eliminate x from the system.
The two lines must be the same (Case 3).
A system that has an infinite number of solutions
is said to be dependent.
© 2010 Pearson Education, Inc. All rights reserved.
1717
Section 8.1, Slide
Solving Systems Using the
Elimination Method
© 2010 Pearson Education, Inc. All rights reserved.
1818
Section 8.1, Slide
Solving a System Using the Elimination Method
The method of eliminating one of the variables to get an
equation, which is easier to solve is known as elimination
method.
19
Solving a System Using the
Substitution Method
20
Solving a System Using the Substitution Method
21
Example 1 – Exploring a System of Equations Using a Table of Values
and Substitution.
Suppose we hired a taxi in New York City and in
Baltimore.
Table 2.12
22
Example 1 – Exploring a System of Equations Using a Table of Values
and Substitution.
(a) Write a system of equations to represent the total costs of hiring a
taxi in New York City and Baltimore as functions of the number of
additional miles traveled beyond the flag drop. (Let m represent the
number of miles traveled beyond the flag drop. Let N(m) represent the
total costs of hiring a taxi in New York and B(m) represent the total
costs of hiring a taxi in Baltimore.) Assume each encounters 5
minutes of wait time during their rides.
23
Example 1 – Exploring a System of Equations Using a Table of Values
and Substitution.
Solution:
N(m) = 2.00m + 4.50
B(m) = 2.20m + 3.80
24
Example 1 – Exploring a System of Equations Using a Table of Values
and Substitution.
(b) Solve the system of equations using
algebra. Verify your solution by graphing the
system or by substituting your solution back
into the original equations. (If an answer
does not exist, enter DNE.)
25
Example 1 – Exploring a System of Equations Using a Table of Values
and Substitution.
Solution:
2.0m + 4.5 = 2.2m +3.8
-2.0m
2.0m
4.5 = .2m + 3.8
-3.8
- 3.8
.7 = .2m
.2
.2
m = 3.5
26
Example 1 – Exploring a System of Equations Using a Table of Values
and Substitution.
Solution:
N(m) = 2.00m + 4.50
B(m) = 2.20m + 3.80
Plug (3.5) in for m into either equation to solve for
your output.
N(3.5) = 2.00(3.5) + 4.50
N(3.5) = 11.50
(3.5, 11.50)
27
Example 1 – Exploring a System of Equations Using a Table of Values
and Substitution.
(c) Explain what your answer in part (b)
means. (3.5, 11.50)
28
Example 1 – Exploring a System of Equations Using a Table of Values
and Substitution.
(c) Solution: The taxi fare in both New York
and Baltimore is $11.50 for a trip 3.5 miles
beyond the flag drop (assuming five minutes
of waiting time).
29
Solving a System Using Graphs
30
Solving a System Using Graphs
One common approach to
solving a system of equations is
to graph all of the equations
simultaneously and find the
point of intersection. We do this
in next Example.
31
Example 2 – Solution
cont’d
Figure 2.16
32
Example 2 – Solution
cont’d
The graphs have been restricted to domains of x 
0 because the number of miles traveled must be
nonnegative to make sense.
The two lines appear to intersect at (2.5, 10).
We estimate this to be the solution of the system;
however, we can use algebra to verify if this is an
exact solution.
33
Breakeven Point
34
Substitution Method: Breakeven Point
Business Applications of Linear Systems
Business analysts are often interested in profit, revenue,
and costs of a company.
In evaluating a business plan, it is important to know at
what sales level revenue and costs are expected to be
equal. This sales level is referred to as the break-even
point and is the point at which the company begins to turn
a profit.
Any business has two types of costs: fixed costs and
variable costs.
35
Solving a System Using the Substitution Method
Fixed costs are those that remain constant
regardless of production levels. For
example, building rent, product research and
development, and advertising are usually
fixed costs.
Variable costs are those that vary with the
level of production. For example, raw
materials and production-line worker wages
are variable costs.
36
Example 5 – Determining a Break-Even Point
An artisan wants to sell her handmade craft angels online.
She estimates her material cost for each angel to be $3.50.
As of January 2007, the online merchant craft mall. com
charged a $14.95 per month fee for a Premier account
featuring up to 25 products.
Comparing her craft to similar crafts on the market, the
artisan estimates she can sell the craft angel for $9.95.
How many angels will she have to sell each month to break
even? At that production level, what will be her production
cost, revenue, and profit?
37
Example 5 – Solution
The cost equation for the craft angels is the sum of the
variable cost, $3.50 per angel, and the fixed cost, $14.95.
Let a be the number of angels sold in a month.
The cost equation is
The revenue equation is
We want to determine when her revenue will equal her
cost.
38
Example 5 – Solution
cont’d
In other words, we want to find the value of a such that
R(a) = C(a). Graphing the two functions simultaneously
results in the graphs shown in Figure 2.18.
Figure 2.18
39
Example 5 – Solution
cont’d
It appears the graphs intersect near (2.3, 23) at the
intersection point, R(a) = C(a). We can find the exact point
of intersection by using the substitution method.
40
Example 5 – Solution
cont’d
We evaluate R(a) at a = 2.318 and determine that
Thus the break-even point is roughly (2.318, 23.06).
In the context of the problem, though, it does not make
sense to talk about 2.318 angels. So we conclude she must
sell 3 angels per month to cover her costs.
41
Example 5 – Solution
cont’d
The cost to produce and advertise 3 angels is
The revenue from the sale of 3 angels is
She will profit $4.40 if she sells 3 angels
42
Solving a System of Three or
More Equations
43
Solving a System of Three or More Equations
For a system of three or more linear
equations to have a solution, all of the
lines must intersect at the same point.
The fact that two lines intersect at a
point (a, b) does not ensure that the
third line will intersect the first two lines
at the same point.
44
Example 11 – Solving a System of Three Equations
Solve the following system of equations.
Solution:
We will find the point of intersection of the first two
equations and then check to see if the point is a solution to
the third equation.
45
Example 11 – Solution
cont’d
The first equation may be written as y = 2x – 5. Substituting
this value in for y in the second equation and solving for x
yields
46
Example 11 – Solution
cont’d
We substitute this value of x into y = 2x – 5 and solve.
The point of intersection of the first two lines is (3, 1). We
will check to see if this point satisfies the third equation.
47
Example 11 – Solution
cont’d
Since the resultant statement is true, the solution to the
system of equations is x = 3 and y = 1. (A false statement
would have shown that the system was inconsistent.) We
confirm the result in Figure 2.21.
Figure 2.21
48