# What is the solution set for this linear-quadratic system of equations? y = x2 − x − 12 y − x − 3 = 0

y - x - 3 = 0 y = x + 3 y = x^2 - x - 12 x + 3 = x^2 - x - 12 x^2 - x - 12 - x - 3 = 0 x^2 - 2x - 15 = 0 (x - 5)(x + 3) = 0 x - 5 = 0 y = x + 3 x = 5 y = 5 + 3 y = 8 x + 3 = 0 y = x + 3 x = -3 y = -3 + 3 y = 0 so ur solutions are : (5,8) and (-3,0)

For this type of solution set, it is easiest to substitute one equation into another before we factor the resulting equation out... y = x² - x - 12 (using this, we already know what y is equal to) y - x - 3 = 0 (substitute the first value of y into this equation and simplify) (x² - x - 12) - x - 3 = 0 (drop the parentheses) x² - x - 12 - x - 3 = 0 (combine like terms) x² - 2x - 15 = 0 (now we can factor -15) (x - 5) (x + 3) = 0 (set each to equal 0) x - 5 = 0 x + 3 = 0 x = 5 x = -3 (now we have the solution to the system)