Solving Systems of Linear
Equations by Graphing
1. Decide whether a given ordered pair is a
solution of a system.
2. Solve linear systems by graphing.
3. Solve special systems by graphing.
4. Identify special systems without graphing.
Solving Systems of Linear Equations
by Graphing
• A system of linear equations, often called a
linear system, consists of two or more linear
equations with the same variables. Examples of
systems include
2x  3y  4
3x  y  5
x  3y  4
 y  4  2x
x  y 1
y  3.
Linear systems
In the system on the right, think of y = 3 as an
equation in two variables by writing it as 0x + y = 3.
Decide whether a given ordered
pair is a solution of a system.
• A solution of a system of a linear equations is
an ordered pair that makes both equations true
at the same time. A solution is said to satisfy
the equation.
EXAMPLE 1
Determining whether an
Ordered Pair Is a Solution
•Decide whether the ordered pair (4,−1) is a
solution of each system.
5 x  6 y  14
2x  5 y  3
 x  y  3
x  y  3
Solution: 5  4   6  1  14
  4    1  3
 4    1  3
2  4   5  1  3
4  1  3
4  1  3
20  6  14
85  3
14  14
33
Yes
3  3
No
3  3
Solve linear systems by graphing.
The set of all ordered pairs that are solutions of a system is its
solution set.
One way to find the solution set of a system of two linear
equations is to graph both equations on the same axes. The graph
of each line shows points whose coordinates satisfy the equation
of that line.
Any intersection point would be on both lines and would
therefore be a solution of both equations. Thus, the coordinates
of any point at which the lines intersect give a solution of the
system. Because the two different straight lines can intersect at
no more then one point, there can never be more than one
solution set for such a system.
Solve linear systems by graphing. (cont’d)
To solve a system by graphing, follow these steps.
Step 1: Graph each equation of the system on the same
coordinate axes.
Step 2: Find the coordinates of the point of intersection of
the graphs if possible. This is the solution of the
system.
Step 3: Check the solution in both of the original equations.
Then write the solution set.
A difficulty with the graphing method is that it may not be possible to
determine from the graph the exact coordinates of the point that
represents the solution, particularly if those coordinates are not integers.
The graphing method does, however, show geometrically how solutions
are found and is useful when approximate answer will do.
Solve a System
by Graphing
EXAMPLE 2
•Solve the system by graphing.
5x  3 y  9
x  2y  7
Solution: {(3,2)}
Solving Special Systems
by Graphing
EXAMPLE 3
•Solve each system by graphing
3x  y  4
6 x  2 y  12
Solution:

2x  5 y  8
4 x  10 y  16
 x, y  2 x  5 y  8
When a system has an infinite number of solutions, either equation of
the system could be used to write the solution set. It’s best to use the
equation (in standard form) with coefficients that are integers
having no common factor (except 1).
Three Cases for Solutions of Systems

The graphs intersect at exactly one point, which gives the (single)
ordered pair solution of the system. The system is consistent and the
equations are independent. See figure (a).
 The graphs are parallel lines, so there is no solution and the solution
set is Ø. The system is inconsistent and the equations are
independent. See figure (b).
 The graphs are the same line. There is an infinite number of
solutions, and the solution set is written in set-builder notation as
{(x,y)|_________}, where one of the equations is written after the |
symbol. The system is consistent and equations are dependent. See
figure (c).
EXAMPLE 4
Identifying the Three Cases by
Using Slopes
Describe each system without graphing. State
the number of solutions.
2x  3y  5
a)
3y  2x  7
2
5
x
3
3
2
7
y 
x
3
3
y 
x  3y  2
6x  y  3
b)
c)
2 x  6 y  4
2 x  y  11
1
2
x
3
3
1
2
y 
x
3
3
y 
y  6 x  3
y  2 x  11
Solution:
a) The equations represent parallel lines. The system has no solution.
b) The equations represent the same line. The system has an infinite
number of solutions.
c) The equations represent lines that are neither parallel nor the same line.
The system has exactly one solution.
EXAMPLE 5
Finding the Solution of a System
by Graphing
•Solve the system by graphing.
5x  3 y  9
x  2y  7
Solution: {(3,2)}
Solving Systems of Linear
Equations by Substitution

Solve linear systems by substitution.

Solve special systems by
substitution.

Solve linear systems with fractions.
Solve linear systems by
substitution.
Graphing to solve a system of equations has a serious drawback.
It is difficult to find an accurate solution, such as
5
1
,



6  , from a graph. One algebraic method for solving a system of
3
equations is the substitution method.
This method is particularly useful for solving systems in which
one equation is already solved, or can be solved quickly, for one of
the variables.
Solve linear systems by substitution. (cont’d)
To solve a system by substitution, follow these steps:

Step 1: Solve one equation for either variable. If one of the
variables has coefficient 1 or −1, choose it, since it
usually makes the substitution method easier.

Step 2: Substitute for that variable in the other equation. The
result should be an equation with just one variable.

Step 3: Solve the equation from Step 2.

Step 4: Substitute the result from Step 3 into the equation
from Step 1 to find the value of the other variable.

Step 5: Check the solution in both of the original equations.
Then write the solution set.
EXAMPLE 1
Using the Substitution Method
•Solve the system by the substitution method.

2 x  7 y  12
x  2 y
Solution:
2  2 y   7 y  12
4 y  7 y  12
3 y 12

3
3
y  4
x  2 y
x  2  4
x 8
8, 4
The solution set found by the substitution method will be the same as
the solution found by graphing. The solution set is the same; only the
method is different.
EXAMPLE 2
Using the Substitution Method
Solve the system by the substitution method.
2 x  7 y  12
x  3 2y
Solution:
2  3  2 y   7 y  12
6  4 y  7 y  12
6  3 y  6  12  6
3 y 18

3
3
x  3  2  6
x  3 12
x  15
15, 6
y  6
Be careful when you write the ordered-pair solution of a system. Even though
we found y first, the x-coordinate is always written first in the ordered pair.
EXAMPLE 3
Using the Substitution Method
•Use substitution to solve the system.
x  1  4 y
2 x  5 y  11
Solution:
x  1  1  4 y  1
x  4 y  1
x  4  1  1
x  4 1
x3
2  4 y 1  5 y  11
8 y  2  5 y  2  11  2
13 y 13

13 13
y  1
 3, 1
EXAMPLE 4
Solving an Inconsistent System
by Substitution
•Use substitution to solve the system.
y  8x  4
16 x  2 y  8
Solution:
16 x  2 8x  4  8
16x 16x  8  8
8  8
Since the statement is false, the solution set is Ø.
It is a common error to give “false” as the solution of an inconsistent
system. The correct response is Ø.
EXAMPLE 5
Solving a System with
Dependent Equations by
Substitution
•Solve the system by the substitution method.
Solution:
x  3 y  7
4 x  12 y  28
x  3 y  3 y  7  3 y
x  7  3 y
4  7  3 y   12 y  28
28  12 y  12 y  28
28  28
Since the statement is true every solution of one equations is also a
solution to the other, so the system has an infinite number of
solutions and the solution set is {(x,y)|x + 3y = −7}.
It is a common error to give “true” as the solution of a system of dependent
equations. Remember to give the solution set in set-builder notation using
the equation in the system that is in standard form with integer coefficients
that have no common factor (except 1).
Using the Substitution Method
with Fractions as Coefficients
EXAMPLE 6
•Solve the system
by
the substitution
method.
1
1
1
x
Solution:
y 
2
3
3
1
x  2 y  2
2
1

2
x

2
y
 
  2  2 
2

3x  2 y  2
3  4  4 y   2 y  2
12  12 y  2 y  12  2  12
10 y
10

10
10
y  1
 0, 1
x  4 y  4 y  4  4 y
x  4  4 y
x  4  4  1
x  4  4
x0
1 
1
1
 6  x  y     6
3 
3
2
Solving Systems of Linear
Equations by Elimination
• Solve linear systems by elimination.
• Multiply when using the elimination method.
• Use an alternative method to find the second value in a
solution.
• Use the elimination method to solve special
systems.
Solve linear systems by elimination.
An algebraic method that depends on the addition property
of equality can also be used to solve systems. Adding the same
quantity to each side of an equation results in equal sums:
If A = B,
then A + C = B + C.
We can take this addition a step further. Adding equal
quantities, rather than the same quantity, to each side of an
equation also results in equal sums:
If A = B, then A + C = B + D.
Using the addition property to solve systems is called the
elimination method. With this method, the idea is to eliminate
one of the variables. To do this, one pair of variable terms in the
two equations must have coefficients that are opposite.
EXAMPLE 1
Using the Elimination Method
Solve the system.
3x  y  7
2x  y  3
Solution:
 3x  y    2 x  y   7  3
5 x 10

5
5
x2
The solution set is
 2, 1.
2  2  y  3
4 y  4  3 4
y  1
A system is not completely solved until values for both x and y are found.
Do not stop after finding the value of only one variable. Remember to write the
solution set as a set containing an ordered pair
Solving a Linear System by Elimination
In general, use the following steps to solve a linear system of
equations by the elimination method.
 Step 1: Write both equations in standard form, Ax + By = C.

Step 2: Transform the equations as needed so that the
coefficients of one pair of variable terms are opposites.
Multiply one or both equations by appropriate numbers so
that the sum of the coefficients of either the x- or y-term is 0.

Step 3: Add the new equations to eliminate a variable. The sum
should be an equation with just one variable.
Step 4: Solve the equation from Step 3 for the remaining variable.
Step 5: Substitute the result from Step 4 into either of the
original equations, and solve for the other variable.



Step 6: Check the solution in both of the original equations.
Then write the solution set.
It does not matter which variable is eliminated first. Choose the one
that is more convenient to work with.
EXAMPLE 2
Solve the system:
Using the Elimination Method
x  2  y
2 x  y  10
Solution:
x  2 y  2  y  y  2
x y 2
 x  y    2x  y   2  10
2 x  y  y  10  y
2 x  y  10
The solution set is
 4, 2.
3 x 12

3
3
x4
4 y4  24
y  2
Multiplying Both Equations
When Using the Elimination
Method
EXAMPLE 3
4 x  5 y  18
3x  2 y  2
Solve the system:
Solution:
 2 4x  5 y   18  2
8 x  10 y  36
15 x  10 y  10
8x 10 y   15x 10 y   36 10
23 x 46

23
23
x  2
The solution set is
53x  2 y   2 5
 2, 2.
3  2  2 y  2
6  2 y  6  2  6
2y 4

2 2
y2
When using the elimination method, remember to multiply both sides of an
equation by the same nonzero number.
EXAMPLE 4
Finding the Second Value by
Using an Alternative Method
Solve the system:
3y  8  4x
6x  9  2 y
Solution:
 2 4x  3 y   8  2
 3 6x  2 y   9  3
+
3 4x  3 y   8 3
 2 6 x  2 y  9  2
8 x  6 y  16
18 x  6 y  27
26x
 11
2 6 x 1 1

26
26
The solution set is
x
+
12 x  9 y  24
12 x  4 y  18
13 y  42
13 y 42

13 13
11
26
 11 42  
 ,   .
 26 16  
42
y
13
Using the Elimination Method
for an Inconsistent System or
Dependent Equations
EXAMPLE 5
Solve each system by the elimination method:
Solution:
3x  y  7
6x  2 y  5
2x  5 y  1
4 x  10 y  2
 23x  y   7  2
 2 2x  5 y   1 2
6 x  2 y  14
6x  2 y  5
0  19
4 x  10 y  2
4 x  10 y  2
00
6x  2 y  5
+
4 x  10 y  2
+
The solution set is
The solution set is
.
 x, y  2 x  5 y  1.
Applications of Linear Systems
1. Solve problems about unknown numbers.
2. Solve problems about quantities and their
costs.
3. Solve problems about mixtures.
4. Solve problems about distance, rate (or
speed), and time.
Applications of Linear Systems
Recall from Section 2.4 the six step method for solving applied
problems. These slightly modified steps allow for two variables
and two equations.
Step 1: Read the problem carefully until you understand what
is given and what is to be found.
Step 2: Assign variables to represent the unknown values,
using diagrams or tables as needed. Write down what
each variable represents.
Step 3: Write two equations using both variables.
Step 4: Solve the system of two equations.
Step 5: State the answer to the problem. Is the answer
reasonable?
Step 6: Check the answer in the words of the original problem.
EXAMPLE 1
Solving a Problem about Two
Unknown Numbers
Two top-grossing Disney movies in 2002 were Lilo and Stitch
and The Santa Clause 2. Together they grossed $284.2 million.
The Santa Clause 2 grossed $7.4 million less than Lilo and Stitch.
How much did each movie gross? (Source: Variety.)
Solution:
Let
x = gross of Lilo and Stitch in millions,
and
y = gross of The Santa Clause 2 in millions.
x  y  284.2
x  7.4  y
 7.4  y   y  7.4  284.2  7.4
Lilo and Stitch grossed 145.8 million dollars and
dollars.
2 y 276.8

2
2
y  138.4
x  7.4 138.4
x  145.8
The Santa Clause 2 grossed 138.4 million
EXAMPLE 2
Solving a Problem about
Quantities and Costs
In 1997 – 1998, the average movie ticket (to the nearest U.S.
dollar) cost $10 in Geneva and $8 in Paris. (Source: Parade,
September 13, 1998.) If a group of 36 people from these two
cities paid $298 for tickets to see The Rookie, how many people
from each city were there?
Number of
Price per Ticket
Total Value
Tickets
(in dollars)
(in dollars)
Paris
x
8
8x
Geneva
y
10
10y
Total
36
XXXXXXXX
298
Solution:
x  y  36
8 x  10 y  298
8  36  y   10 y  298
288  8 y  10 y  288  298  288
2 y 10

2
2
y5
x  36  5
x  31
There were 5 people from Geneva, and 31 people from Paris that went to see The Rookie.
EXAMPLE 3
Solving a Mixture Problem
Involving Percent
How many liters of a 25% alcohol solution must be mixed with a 12% solution to get 13 L of a
15% solution?
Solution:
x  y  13
.12 x  .25 y  1.95
100.12x  .25 y   1.95 100
12 13  y   25 y  195
156  12 y  25 y  156  195  156
Liters of
Percent (as
Liters of
Solution
a decimal)
pure alcohol
x
.12
.12x
y
.25
.25y
13
.15
1.95
13 y 39

13 13
y3
x  13  3
x  10
To make 13 L of a 15% solution, 3 L of 25% solution, and 10 L of 12% solution must be used.
Recall working these mixture problems! We used one variable.
EXAMPLE 4
Solving a Problem about
Distance, Rate, and Time
In one hour, Abby can row 2 mi against the current or 10 mi with the current. Find the speed of
the current and Abby’s speed in still water.
Solution:
Let
x = Abby’s speed in still water in mph,
and
y = the water speed of the current in mph.
x  y  10
x y 2
 2  y   y  2  10  2
2y 8

2 2
y4
Abby’s speed in still water is 6 mph, and the speed of the current is 4 mph.
x  24
x6
SYSTEMS OF LINEAR
INEQUALITIES
Solving Linear Systems of Inequalities by
Graphing
Solving Systems of Linear
Inequalities
1. We show the solution to a system of
linear inequalities by graphing them.
a) This process is easier if we put the
inequalities into Slope-Intercept
Form, y = mx + b.
Solving Systems of Linear
Inequalities
2. Graph the line using the y-intercept &
slope.
a) If the inequality is < or >, make the
lines dotted.
b) If the inequality is < or >, make the
lines solid.
Solving Systems of Linear
Inequalities
3. The solution also includes points not
on the line, so you need to shade the
region of the graph:
above the line for ‘y >’ or ‘y ’.
b) below the line for ‘y <’ or ‘y ≤’.
a)
Solving Systems of Linear
Inequalities
Example:
a: 3x + 4y > - 4
b:
x + 2y < 2
Put in Slope-Intercept Form:
a)3x  4 y  4
b) x  2 y  2
4 y  3x  4
2 y   x  2
3
 y   x  1
4
1
 y   x  1
2
Solving Systems of Linear
Inequalities
Example, continued:
3
a : y   x  1
4
1
b : y   x  1
2
Graph each line, make dotted or solid
and shade the correct area.
a:
dotted
shade above
b:
dotted
shade below
Solving Systems of Linear
Inequalities
a: 3x + 4y > - 4
3
a : y   x  1
4
Solving Systems of Linear
Inequalities
a: 3x + 4y > - 4
b: x + 2y < 2
3
a : y   x  1
4
1
b : y   x  1
2
Solving Systems of Linear
Inequalities
a: 3x + 4y > - 4
b: x + 2y < 2
The area between the
green arrows is the
region of overlap and
thus the solution.
Solving Systems of Three Linear
Equations in Three Variables
The Elimination Method
Solutions of a system with 3
equations
The solution to a system of three linear
equations in three variables is an ordered
triple.
(x, y, z)
The solution must be a solution of all 3
equations.
Is (–3, 2, 4) a solution of this
system?
3x + 2y + 4z = 11
2x – y + 3z = 4
5x – 3y + 5z = –1
3(–3) + 2(2) + 4(4) = 11
2(–3) – 2 + 3(4) = 4 P
5(–3) – 3(2) + 5(4) = –1
P
P
Yes, it is a solution to the system because it is a
solution to all 3 equations.
Use elimination to solve the following
system of equations.
x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
Step 1
Rewrite the system as two smaller systems,
each containing two of the three equations.
x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
x – 3y + 6z = 21
x – 3y + 6z = 21
3x + 2y – 5z = –30 2x – 5y + 2z = –6
Step 2
Eliminate THE SAME variable in each of
the two smaller systems.
Any variable will work, but sometimes one
may be a bit easier to eliminate.
I choose x for this system.
(x – 3y + 6z = 21) (–3)
3x + 2y – 5z = –30
(x – 3y + 6z = 21) (–2)
2x – 5y + 2z = –6
–3x + 9y – 18z = –63
3x + 2y – 5z = –30
–2x + 6y – 12z = –42
2x – 5y + 2z = –6
11y – 23z = –93
y – 10z = –48
Step 3
Write the resulting equations in two
variables together as a system of equations.
Solve the system for the two remaining
variables.
11y – 23z = –93
y – 10z = –48 (–11)
11y – 23z = –93
–11y + 110z = 528
87z = 435
z=5
y – 10(5) = –48
y – 50 = –48
y=2
Step 4
Substitute the value of the variables from the
system of two equations in one of the
ORIGINAL equations with three variables.
x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
I choose the first equation.
x – 3(2) + 6(5) = 21
x – 6 + 30 = 21
x + 24 = 21
x = –3
Step 5
CHECK the solution in ALL 3 of the
original equations.
Write the solution as an ordered triple.
x – 3y + 6z = 21
3x + 2y – 5z = –30
2x – 5y + 2z = –6
P
–3 – 3(2) + 6(5) = 21
3(–3) + 2(2) – 5(5) = –30
2(–3) – 5(2) + 2(5) = –6
The solution is (–3, 2, 5).
P
P
It is very helpful to neatly organize your
work on your paper in the following manner.
(x, y, z)
Solve the system.
1. x+3y-z=-11
2x+y+z=1
z’s are easy to cancel!
3x+4y=-10
2. 2x+y+z=1
5x-2y+3z=21
Must cancel z’s again!
-6x-3y-3z=-3
5x-2y+3z=21
-x-5y=18
2(2)+(-4)+z=1
z=1
4-4+z=1
x+3y-z=-11
2x+y+z=1
5x-2y+3z=21
3. 3x+4y=-10
-x-5y=18
Solve for x & y.
3x+4y=-10
-3x-15y+54
-11y=44
y=- 4
3x+4(-4)=-10
x=2
(2, - 4, 1)
-x+2y+z=3
2x+2y+z=5
4x+4y+2z=6
Solve the system.
1. -x+2y+z=3
2x+2y+z=5
z’s are easy to cancel!
-x+2y+z=3
-2x-2y-z=-5
-3x=-2
x=2/3
2.
2x+2y+z=5
4x+4y+2z=6
Cancel z’s again.
-4x-4y-2z=-10
4x+4y+2z=6
0=- 4
Doesn’t make sense!
No solution
Solve the system.
1. -2x+4y+z=1
3x-3y-z=2
z’s are easy to cancel!
x+y=3
2. 3x-3y-z=2
5x-y-z=8
Cancel z’s again.
-3x+3y+z=-2
5x-y-z=8
2x+2y=6
-2x+4y+z=1
3x-3y-z=2
5x-y-z=8
3. x+y=3
2x+2y=6
Cancel the x’s.
-2x-2y=-6
2x+2y=6
0=0
This is true.
¸ many solutions
Solve:
x – 6y – 2z = –8
–x + 5y + 3z = 2
3x – 2y – 4z = 18
Answer:
x – 6y – 2z = –8
–x + 5y + 3z = 2
3x – 2y – 4z = 18
(4, 3, –3)
Solve:
–5x + 3y + z = –15
10x + 2y + 8z = 18
15x + 5y + 7z = 9
Answer:
–5x + 3y + z = –15
10x + 2y + 8z = 18
15x + 5y + 7z = 9
(1, –4, 2)
Application
Courtney has a total of 256 points on three
Algebra tests. His score on the first test
exceeds his score on the second by 6 points.
His total score before taking the third test was
164 points. What were Courtney’s test scores
on the three tests?
Explore
Problems like this one can be solved using a
system of equations in three variables.
Solving these systems is very similar to solving
systems of equations in two variables. Try
solving the problem
Let f = Courtney’s score on the first test
Let s = Courtney’s score on the second test
Let t = Courtney’s score on the third test.
Plan
Write the system of equations from the
information given.
f + s + t = 256
f–s=6
f + s = 164
The total of the scores is 256.
The difference between the 1st and 2nd is 6 points.
The total before taking the third test is the sum of the first and
second tests..
Solve
Now solve. First use elimination on the last two
equations to solve for f.
f–s=6
f + s = 164
2f = 170
The first test score is 85.
f = 85
Solve
Then substitute 85 for f in one of the original
equations to solve for s.
f + s = 164
85 + s = 164
s = 79 The second test score is 79.
Solve
Next substitute 85 for f and 79 for s in f + s + t =
256.
f + s + t = 256
85 + 79 + t = 256
164 + t = 256
t = 92 The third test score is 92.
Courtney’s test scores were 85, 79, and 92.
Examine
Now check your results against the original problem.
Is the total number of points on the three tests 256
points?
85 + 79 + 92 = 256 ✔
Is one test score 6 more than another test score?
79 + 6 = 85 ✔
Do two of the tests total 164 points?
85 + 79 =164 ✔
Our answers are correct.
Solutions?
You know that a system of two linear equations
doesn’t necessarily have a solution that is a
unique ordered pair. Similarly, a system of
three linear equations in three variables
doesn’t always have a solution that is a unique
ordered triple.
Graphs
The graph of each equation in a system of three
linear equations in three variables is a plane.
Depending on the constraints involved, one
of the following possibilities occurs.
Graphs
1.
The three planes intersect
at one point. So the
system has a unique
solution.
2. The three planes intersect
in a line. There are an infinite
number of solutions to the
system.
Graphs
3. Each of the diagrams below shows three planes that
have no points in common. These systems of
equations have no solutions.
Solve this system of equations
x  2y  z  9
3 y  z  1
3z  12
Answer:
x  2y  z  9
3 y  z  1
3z  12
Solve the third equation, 3z = 12
3z = 12
z=4
Substitute 4 for z in the second
equation 3y – z = -1 to find y.
3y – (4) = -1
3y = 3
y=1
Substitute 4 for z and 1 for y in the
first equation, x + 2y + z = 9 to find x.
x + 2y + z = 9
x + 2(1) + 4 = 9
x+6=9
x = 3 Solution is (3, 1, 4)
Check:
1st 3 + 2(1) +4 = 9 ✔
2nd 3(1) -4 = 1 ✔
3rd 3(4) = 12 ✔
Solve this system of equations
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Answer:
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Set the first two equations together
and multiply the first times 2.
2(2x – y + z = 3)
4x – 2y +2z = 6
x + 3y -2z = 11
5x + y = 17
Set the next two equations together
and multiply the first times 2.
2(x + 3y – 2z = 11)
2x + 6y – 4z = 22
3x - 2y + 4z = 1
5x + 4y = 23
Next take the two equations that
only have x and y in them and put
them together. Multiply the first
times -1 to change the signs.
Answer:
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Next take the two equations that only
have x and y in them and put them
together. Multiply the first times -1 to
change the signs.
-1(5x + y = 17)
-5x - y = -17
5x + 4y = 23
3y = 6
y=2
Now you have y = 2. Substitute y into
one of the equations that only has an x
and y in it.
5x + y = 17
5x + 2 = 17
5x = 15
x=3
Now you have x and y. Substitute values
back into one of the equations that you
started with.
2x – y + z = 3
2(3) - 2 + z = 3
6–2+z=3
4+z=3
z = -1
Check your work!!!
2x  y  z  3
x  3 y  2 z  11
3x  2 y  4 z  1
Solution is (3, 2, -1)
Check:
1st 2x – y + z =
2(3) – 2 – 1 = 3 ✔
2nd x + 3y – 2z = 11
3 + 3(2) -2(-1) = 11 ✔
3rd 3x – 2y + 4z
3(3) – 2(2) + 4(-1) = 1 ✔
Competition Problems
Find the solution to the linear
system:
-x + 5y = -23
4x - 2y = 20
Answer:
x=3
y = -4
If A x B =8, B x C = 28, and
A x C = 14,
find A x B x C
Answer:
56
A polygon is formed by the intersections of
y ≥ 3x – 3
y≤3
x ≥ –2 and
3x + 2y ≥ –6
What is the area of the polygon?
Answer:
15
Given the system
y  0 .5 x
y   0 .5 x  2
y x2
graph the figure formed by the solution of
the system?
Answer:
scalene triangle
Line m has the equation 3y = 6 + y. Line
n is perpendicular to line m and contains
the point (4, –2). When
graphed, what is the point of
intersection for lines m and n?
Answer:
(4,3)
Find the area bounded by
x = –1, x = 5, y = 0 and y = –2x + 13.
Answer:
54