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.

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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 + 4x 7x + 2y – 2y = 4 = –18 = –14

The y-terms have opposite coefficients. Add the equations to eliminate y.

x = –2

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 –3 (3 (2

x x

+ 5 + 3

y y

) = ) = 2 ( –3 –16) ( –9) 6

x

+ 10

y

= –32 –6

x

– 9

y

= 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.

3

x

+ 5 ( –5) = –16 3

x

– 25 = –16 3

x

= 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. 3 (3) 3

x

+ 5

y

= –16 + 5 ( –5) –16 –16 –16  2

x

+ 3

y

= –9 2 (3) + 3 ( –5) –9 –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 – 12x + 7y – 7y = –25 = 19 –8x = –6

The y-terms have opposite coefficients. Add the equations to eliminate y.

x =

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 (5

x

– 3

y

) = –8 (42) 5 (8

x

+ 5

y

) = 5 (28) –40

x

+ 24

y

= –336 40

x

+ 25

y

49

y y

= 140 = –196

Add the equations.

= –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.

5

x

– 3 ( –4) = 42 5

x

+ 12 = 42 5

x

= 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. 5

x

– 3

y

= 42 5 (6) – 3 ( –4) 42 42 42  8

x

+ 5

y

= 28 8 (6) + 5 ( –4) 28 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 2 – 6x + 6x = –18 2 = –18

Substitute (1 –3x) for y in the second equation.

Distribute.

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

Solve the second equation.

6x + 3 (–6 – 2x) = –12

Substitute ( –6 – 2x) for y in the first equation.

6x –18 – 6x = –12 –18 = –12

x

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

x

plus + amount of bacon mix

y

equals = 60.

60 Write another equation based on the amount of protein: Protein of beef mix 0.18x plus + protein of bacon mix 0.09y equals = protein in mixture.

0.15(60)

Example 4 Continued

Solve the system.

x + y = 60 0.18x +0.09y = 9 x + y = 60 y = 60 – x 0.18x + 0.09

(60 – x) = 9 0.18x + 5.4 – 0.09x = 9 0.09x = 3.6

x = 40

First equation Solve the first equation for y.

Substitute (60 – x) for y.

Distribute.

Simplify.

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

x

+

y

equals = 50.

50 Write another equation based on cost of the beans: Cost of Sumatra beans 5x plus + cost of 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 y = 50 – x 5x + 13 (50 – x) = 500 5x + 650 – 13x = 500 –8x = –150 x = 18.75

First equation 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.