Question

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

y = x2 − x − 12
y − x − 3 = 0

Answer 1

Solving the second for y:

y=x+3

And since any solution will satisfy y=y we can say:

x^2-x-12=x+3 subtract x+3 from both sides

x^2-2x-15=0 now factor

x^2+3x-5x-15-0

x(x+3)-5(x+3)=0

(x+3)(x-5)=0

So there are two solution which occur when x=-3 and 5. Using y=x+3, you can find the corresponding y values...

y(-3)=-3+3=0 and y(5)=5+3=8

So the two solutions are the points (-3,0) and (5,8)