Solving Systems of Equations

Using Elimination

Solving Systems by Elimination

Another method for solving systems of equations is elimination. Like substitution, the goal of elimination is to get one equation that has only one variable.

Elimination Using Addition

Example 1 Elimination Using Addition

Use elimination to solve each system of equations.

-4x + 3y = 17

4x + y = 3

Since the coefficients of the x-terms, -4 and 4, are additive inverses, you can eliminate the x-terms by adding the equations.

-4x + 3y = 17 Write the equations in column form and add.

(+)4x + y = 3

4y = 20 Notice the x variable is eliminated.

4y= 20 Divide each side by 4.

y = 5 Simplify.

Now substitute 5 for y in either equation to find the value of x.

4x + y = 3 Second equation

4x + 5 = 3 Replace y with 5.

4x + 5 – 5 = 3 – 5 Subtract 5 from each side.

4x = -2 Simplify.

4x= -2 Divide each side by 4.

x = - (1/2) Simplify.

The solution is (-1/2, 5)

Review

Elimination Using Subtraction

Hint: same term, you subtract

Example 2 Elimination Using Subtraction

Use elimination to solve the system of equations.

7a + 3b = 3

2a + 3b = 18

Since the coefficients of the b-terms, 3 and 3, are the same, you can eliminate the b-terms by subtracting the equations.

7a + 3b = 3 Write the equations in column form and subtract.

(-) 2a + 3b = 18

5a = -15 The variable b is eliminated.

5a= -15 Divide each side by 5.

a = -3 Simplify.

Now substitute –3 for a in either equation to find the value of b.

2a + 3b = 18 Second equation

2(-3) + 3b = 18 a = -3

-6 + 3b = 18 Simplify.

3b = 24 Add 6 to each side and simplify.

3b= 24 Divide each side by 3.

b = 8

The solution is (-3, 8).

Using Multiplication

Example 1 Multiply One Equation to Eliminate

Use elimination to solve the system of equations.

x + 3y = -4

x + 2y = 9

Multiply the first equation by –3 so the coefficients of the x-terms are additive inverses. Then add the equations.

x + 3y = -4 Multiply by –3. -x – 9y = 12

x + 2y = 9 (+) x + 2y = 9

-7y = 21 Add the equations.

= Divide each side by –7.

y = -3

Now substitute –3 for y in either equation to find the value of x.

x + 2y = 9 Second equation

x + 2(-3) = 9 y = -3

x – 6 = 9 Simplify.

x = 15 Add 6 to each side and simplify.

The solution is (15, -3).

Using Multiplication

Example 2 Multiply Both Equations to Eliminate

Use elimination to solve the system of equations.

5x – 7y = -2

-4x + 6y = 4

Method 1 Eliminate x.

5x – 7y = -2 Multiply by 4. 20x – 28y = -8

-4x + 6y = 4 Multiply by 5. (+) -20x + 30y = 20

2y = 12 Add the equations.

2y= 12 Divide each side by 2.

y = 6 Simplify.

Now substitute 6 for y in either equation to find the value of x.

5x – 7y = -2 First equation

5x – 7(6) = -2 y = 6

5x – 42 = -2 Simplify.

5x – 42 + 42 = -2 + 42 Add 42 to each side.

5x = 40 Simplify.

x = 8 Divide each side by 5 and simplify.

The solution is (8, 6).