1. Objective
The student will be able to: solve systems of equations using elimination with multiplication.

2. Solving Systems of Equations
So far, we have solved systems using graphing, substitution, and elimination.
This goes one step further and show how to use ELIMINATION with multiplication.
What happens when the coefficients are not the same?
We multiply the equations to make them the same! You’ll see…

3. Solving a system of equations by elimination using multiplication.
Step 1: Put the equations in Standard Form.
Standard Form: Ax + By = C
Step 2: Determine which variable to eliminate.
Look for variables that have the
same coefficient.
Step 3: Multiply the equations and solve.
Solve for the variable.
Step 4: Plug back in to find the other variable.
Substitute the value of the variable
into the equation.
Step 5: Check your solution.
Substitute your ordered pair into
BOTH equations.

4. 1) Solve the system using elimination.
2x + 2y = 6
3x – y = 5
Multiply the bottom
equation by 2
2x + 2y = 6
(2)(3x – y = 5)
2x + 2y = 6
(+) 6x – 2y = 10
8x = 16
x = 2
2(2) + 2y = 6
4 + 2y = 6
2y = 2
y = 1
(2, 1)

5. 2) Solve the system using elimination.
x + 4y = 7 4x – 3y = 9
Multiply the top equation by -4
(-4)(x + 4y = 7)
4x – 3y = 9
-4x – 16y = -28
(+) 4x – 3y = 9
-19y = -19
y = 1
4x – 3y = 9
4x – 3(1) = 9
4x – 3 = 9
4x = 12
x = 3
(3, 1)

6. Which variable is easier to eliminate?
3x + y = 4
4x + 4y = 6
y is easier to eliminate.
Multiply the 1st equation by (-4) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

7. 3) Solve the system using elimination.
3x + 4y = -1 4x – 3y = 7
Multiply both equations
(3)(3x + 4y = -1)
(4)(4x – 3y = 7)
9x + 12y = -3
(+) 16x – 12y = 28
25x = 25
x = 1
3(1) + 4y = -1
3 + 4y = -1
4y = -4
y = -1
(1, -1)

8. 4) What is the best number to multiply the top equation by to eliminate the x’s?
3x + y = 4
6x + 4y = 6
Multiply the 1st equation by (-2)
(-2)(3x + y = 4)  -6x -2y = -8
6x + 4y = 6
2y = -2
y = 1
6x + 4y = 6
6x + 4(1) = 6
6x = 2
x = 2/6
x = 1/3
(1/3, 1) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

9. 5) Solve using elimination.
2x – 3y = 1
x + 2y = -3
Multiply 2nd equation by (-2) to eliminate the x’s.
(-2)(x + 2y = -3)  -2x – 4y = 6
1st equation  2x – 3y = 1
-7y = 7
y = -1
2x – 3y = 1
2x – 3(-1) = 1
2x + 3 = 1
2x = -2
x = -1
(-1, -1)

10. 5) Solve using elimination.
2(4x + y = 6 ) -8x – 2y = 13 8x + 2y = 12 0 = 25
Add down to eliminate x.
But look what happens, y is eliminated too.
We now have a false statement, thus the system has no solution, it is inconsistent.

11. 6) Elimination Solve the system using elimination. 5x – 2y = –15
(4)
Since neither variable will drop out if the equations are added together.
Multiply one or both of
the equations by a constant to make one of the variables have the same number with opposite signs.
The best choice is to multiply the top equation by 4 since only one equation would have to be multiplied.
Also, the signs on the y-terms are
already opposites.

12. 6) Elimination 3x + 8y = 37 (second equation) 3(–1) + 8y = 37
The solution is (–1, 5)
To find the second variable, substitute in any equation that contains two variables.

13. 7) Elimination Solve the system using elimination. 4x + 3y = 8
(5)
For this system, we must multiply both equations by a different constant in order to make one of the variables “drop out.”
(3)
It would work to multiply the top equation by –3 and the bottom equation by 4 OR to multiply the top equation by 5 and the bottom equation by 3.

14. 7) Elimination 4x + 3y = 8 4(–1) + 3y = 8 –4 + 3y = 8 3y = 12 y = 4
The solution is (–1, 4)

15. 8) Elimination Method Solve the system. (1) (2)
To eliminate x, multiply equation (1) by –2 and
equation (2) by 3 and add the resulting equations.
(1)
(2)

16. 8) Elimination Method The solution is (3, 2)
Substitute 2 for y in (1) or (2).
The solution is (3, 2)

17. 9) Solving an Inconsistent System
Solve the system
Eliminate x by multiplying (1) by 2 and
adding the result to (2).
Solution set is .
(1)
(2)
Inconsistent System

18. 10) Solving a System with Dependent Equations
Solve the system.
Eliminate x by multiplying (1) by 2 and adding the result to (2).
Each equation is a solution of the other.
Infinite solutions.
(1)
(2)