Question

What is the solution set for this linear-quadratic system of equations?
y = x^2 − x − 12
y − x − 3 = 0

Answer 1

@welshfella @acxbox22

Answer 2

{(-3, 0), (0, 3)}
{(-3, 0), (4, 0)}
{(-3, 0), (5, 8)}
{(4, 0), (0, 3)}

Answer 3

first solve for y on the bottome equation
y-x-3=0
y=x+3
now we can plug in (x+3) for y on the top equation

Answer 4

x+3=x^2-x-12
subtracting x+3 on the left side we get
0=x^2-2x-15
now factor that and solve for x

Answer 5

So its c.

Answer 6

0=(x-5)(x+3)
x=5,-3
now plug those two values into wither equation to get their respective y values