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

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?

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)

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)

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

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)

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

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)

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.

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

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)

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