Question

What is the solution set for this linear-quadratic system of equations?

Answer 1

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)

Answer 2

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)