1. Objective
I CAN solve systems of equations using elimination with multiplication.

2. Solving Systems of Equations
On Tuesday we learned that we have to have an opposite to eliminate a variable.
2x + 5y = 10
3x – 5y = 12
What happens when the coefficients are not the same?

3. 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)

4. 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)

5. 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

6. 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)

7. 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

8. 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)

9. 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.

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

11. 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)

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