1. Systems of Equations Elimination

2. This method is a little harder than substitution because you have to think a bit! In elimination, you will be “eliminating” a variable to solve the problems. Which variable is determined by what the problem looks like or what you feel like getting rid of first.

3. Elimination Solve using elimination: 2x – y = 5 -x + y = - 2 2x – y = 5 -x + y = - 2 x = 3 2(3) – y = 5 6 – y = 5 - y = - 1 y = 1 (3, 1) 1.Add the equations together to get rid of a variable. 2.Plug the variable back into one of the equations to get the other variable.

4. Elimination That was an easy one. Sometimes you have to multiply to get rid of a variable. 2x – y = -6 4x - y = 6 (2x – y = -6) -1 - 2x + y = 6 4x – y = 6 2x = 12 x = 6 2(6) – y = - 6 12 – y = - 6 - y = - 18 y = 18 (6, 18) 1.You have to multiply one equation to get rid of a variable. 2.Add the two equations and then solve for the variable. 3.Plug the variable back in and solve for the other variable.

5. Elimination 2x – 3y = 9 4x – 4y = 2 (2x – 3y = 9) -2 -4x + 6y = -18 4x – 4y = 2 2y = - 16 y = - 8 2x – 3(-8) = 9 2x + 24 = 9 2x = -15 x = -15/2 (-15/2, -8)

6. Elimination Try a couple on your own! 2x + 3y = 1x – 5y = 2x – y = 14 4x – 3y = -7-10x + 50y = -20-x + y = 9 (-1, 1) Infinitely ManyNo Solution