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You can also solve systems of equations with the
elimination method. With elimination, you get rid of
one of the variables by adding or subtracting
equations. You may have to multiply one or both
equations by a number to create variable terms that
can be eliminated.
Reading Math
The elimination method is sometimes called the
addition method or linear combination.
Example 2A: Solving Linear Systems by Elimination
Use elimination to solve the system of equations.
3x + 2y = 4
4x – 2y = –18
Step 1 Find the value of one variable.
3x + 2y = 4
+ 4x – 2y = –18
7x
= –14
x = –2
The y-terms have opposite coefficients.
Add the equations to eliminate y.
First part of the solution
Example 2A Continued
Step 2 Substitute the x-value into one of the
original equations to solve for y.
3(–2) + 2y = 4
2y = 10
y=5
Second part of the solution
The solution to the system is (–2, 5).
Example 2B: Solving Linear Systems by Elimination
Use elimination to solve the system of equations.
3x + 5y = –16
2x + 3y = –9
Step 1 To eliminate x, multiply both sides of the
first equation by 2 and both sides of the second
equation by –3.
2(3x + 5y) = 2(–16)
–3(2x + 3y) = –3(–9)
6x + 10y = –32
–6x – 9y = 27
y = –5
Add the equations.
First part of the solution
Example 2B Continued
Step 2 Substitute the y-value into one of the
original equations to solve for x.
3x + 5(–5) = –16
3x – 25 = –16
3x = 9
x =3
Second part of the solution
The solution for the system is (3, –5).
Example 2B: Solving Linear Systems by Elimination
Check Substitute 3 for x and –5 for y in each equation.
3x + 5y = –16
2x + 3y = –9
3(3) + 5(–5) –16
2(3) + 3(–5) –9
–16
–16 
–9
–9 
Check It Out! Example 2a
Use elimination to solve the system of equations.
4x + 7y = –25
–12x –7y = 19
Step 1 Find the value of one variable.
4x + 7y = –25
– 12x – 7y = 19
–8x
= –6
x=
The y-terms have opposite coefficients.
Add the equations to eliminate y.
First part of the solution
Check It Out! Example 2a Continued
Step 2 Substitute the x-value into one of the
original equations to solve for y.
4(
) + 7y = –25
3 + 7y = –25
7y = –28
y = –4
Second part of the solution
The solution to the system is (
, –4).
Check It Out! Example 2b
Use elimination to solve the system of equations.
5x – 3y = 42
8x + 5y = 28
Step 1 To eliminate x, multiply both sides of the
first equation by –8 and both sides of the second
equation by 5.
–8(5x – 3y) = –8(42)
5(8x + 5y) = 5(28)
–40x + 24y = –336
40x + 25y = 140 Add the equations.
49y = –196
y = –4
First part of
the solution
Check It Out! Example 2b
Step 2 Substitute the y-value into one of the
original equations to solve for x.
5x – 3(–4) = 42
5x + 12 = 42
5x = 30
x =6
Second part of the solution
The solution for the system is (6,–4).
Check It Out! Example 2b
Check Substitute 6 for x and –4 for y in each equation.
5x – 3y = 42
8x + 5y = 28
5(6) – 3(–4) 42
8(6) + 5(–4) 28
42
42 
28
28 
In Lesson 3–1, you learned that systems may
have infinitely many or no solutions. When you
try to solve these systems algebraically, the
result will be an identity or a contradiction.
Remember!
An identity, such as 0 = 0, is always true and
indicates infinitely many solutions. A
contradiction, such as 1 = 3, is never true and
indicates no solution.
Example 3: Solving Systems with Infinitely Many or
No Solutions
Classify the system and determine the number of
solutions.
3x + y = 1
2y + 6x = –18
Because isolating y is straightforward, use substitution.
3x + y = 1
y = 1 –3x Solve the first equation for y.
2(1 – 3x) + 6x = –18 Substitute (1–3x) for y in the second equation.
2 – 6x + 6x = –18 Distribute.
2 = –18 x Simplify.
Because 2 is never equal to –18, the equation is a
contradiction. Therefore, the system is inconsistent and
has no solution.
Check It Out! Example 3a
Classify the system and determine the number of
solutions.
56x + 8y = –32
7x + y = –4
Because isolating y is straightforward, use substitution.
7x + y = –4
y = –4 – 7x
Solve the second equation for y.
56x + 8(–4 – 7x) = –32
56x – 32 – 56x = –32
Substitute (–4 –7x) for y in the first
equation.
Distribute.
–32 = –32  Simplify.
Because –32 is equal to –32, the equation is an identity.
The system is consistent, dependent and has infinite
number of solutions.
Check It Out! Example 3b
Classify the system and determine the number of
solutions.
6x + 3y = –12
2x + y = –6
Because isolating y is straightforward, use substitution.
2x + y = –6
y = –6 – 2x
6x + 3(–6 – 2x)= –12
6x –18 – 6x = –12
–18 = –12 x
Solve the second equation.
Substitute (–6 – 2x) for y in the first
equation.
Distribute.
Simplify.
Because –18 is never equal to –12, the equation is a contradiction.
Therefore, the system is inconsistent and has no solutions.
Example 4: Zoology Application
A veterinarian needs 60 pounds of dog food
that is 15% protein. He will combine a beef
mix that is 18% protein with a bacon mix that
is 9% protein. How many pounds of each does
he need to make the 15% protein mixture?
Let x present the amount of beef mix in the mixture.
Let y present the amount of bacon mix in the mixture.
Example 4 Continued
Write one equation based on the amount of dog food:
Amount of
beef mix
plus
amount of
bacon mix
equals
60.
+
y
=
60
x
Write another equation based on the amount of protein:
Protein of
beef mix
0.18x
plus
protein of
bacon mix
equals
protein in
mixture.
+
0.09y
=
0.15(60)
Example 4 Continued
Solve the system.
x + y = 60
0.18x +0.09y = 9
x + y = 60
First equation
y = 60 – x
Solve the first equation for y.
0.18x + 0.09(60 – x) = 9
Substitute (60 – x) for y.
0.18x + 5.4 – 0.09x = 9 Distribute.
0.09x = 3.6 Simplify.
x = 40
Example 4 Continued
Substitute x into one of the original equations to solve
for y.
40 + y = 60
y = 20
Substitute the value of x into
one equation.
Solve for y.
The mixture will contain 40 lb of the beef mix and 20 lb
of the bacon mix.
Check It Out! Example 4
A coffee blend contains Sumatra beans which
cost $5/lb, and Kona beans, which cost
$13/lb. If the blend costs $10/lb, how much
of each type of coffee is in 50 lb of the blend?
Let x represent the amount of the Sumatra beans
in the blend.
Let y represent the amount of the Kona beans in
the blend.
Check It Out! Example 4 Continued
Write one equation based on the amount of each bean:
Amount of
Sumatra beans
plus
x
+
amount of Kona
beans
equals
=
y
50.
50
Write another equation based on cost of the beans:
Cost of Sumatra
beans
5x
cost of
plus
Kona beans
+
13y
equals
cost of
beans.
=
10(50)
Check It Out! Example 4 Continued
Solve the system.
x + y = 50
5x + 13y = 500
x + y = 50
First equation
y = 50 – x
5x + 13(50 – x) = 500
5x + 650 – 13x = 500
–8x = –150
x = 18.75
Solve the first equation for y.
Substitute (50 – x) for y.
Distribute.
Simplify.
Check It Out! Example 4 Continued
Substitute x into one of the original equations to solve
for y.
18.75 + y = 50
y = 31.25
Substitute the value of x into
one equation.
Solve for y.
The mixture will contain 18.75 lb of the Sumatra beans
and 31.25 lb of the Kona beans.