Solving Quadratics by Substitution y 2 – x – 6 = 0 x + y = 0 Solve for x or y Plug in the expression for y. Simplify and combine like terms Solve the Quadratic by factoring.(If you can’t factor use quadratic formula or completing the square) Plug your answers back into one of the original problems. Solutions are point of intersection. y = -x (-x) 2 – x – 6 = 0 x 2 – x – 6 = 0 (x +2)(x – 3)= 0 x = -2 x = 3 y = - (-2) y = -(3) y = 2 y = -3 (-2,2) (3,-3)

Solving Quadratics by Substitution Solve for x or y Plug in the expression for y. Simplify and combine like terms Solve the Quadratic by factoring.(If you can’t factor use quadratic formula or completing the square) Plug your answers back into one of the original problems. Solutions are point of intersection. x 2 +6x + 4y – 3 = 0 y + 3x + 1 = 0

Solving Quadratics by Substitution y = x 2 + 2x – 3 y = -x 2 – 2x + 3 Solve for x or y Since they are both already = y you can set them = to each other. Simplify and combine like terms Solve the Quadratic by factoring.(In this problem you can divivide by 2 first) Plug your answers back into one of the original problems. Solutions are point of intersection. x 2 + 2x – 3 = -x 2 – 2x + 3 2x 2 + 4x – 6 = 0 x 2 + 2x - 3 = 0 (x – 1)(x + 3)= 0 x = 1 x = -3 y = (1) – 3 y = (-3) 2 + 2(-3) – 3 y = y = 9 – 6 – 3 y = 0 y = -4 (1,0) (-3,0)

Practice Problems

Assignment #1 Solving systems with quadratics ws.