Question

What is the solution of the linear-quadratic system of equation?

Answer 1

y = x^2 + 5x - 3
and
y - x = 2


you can substitute, i would solve the second one for y, and put that into the first one

y = 2 + x

use that for y in the first equation

2 + x = x^2 + 5x - 3

Answer 2

@DanJS I assume we'd organize it so it'd be like x-5x-x^2=-3-3 ? I understand how to simplify the polynomials and the variables like x, but I'm not sure what to do with x^2..

Answer 3

take 2 off both sides and take x off both sides to get

0 = x^2 + 4x - 5

Answer 4

you can use the quadratic formula to solve for x all the time for these, but this one factors nicely

0 = (x + 5)*(x - 1)

Answer 5

@DanJS Oh! Is that the answer?

Answer 6

That is true if either (x+5) or the (x-1) are zero, so

x + 5 =0
or
x - 1 = 0


x = -5
or
x=1

Answer 7

the line intersects the parabola twice, when x=-5 and x=1, the y values to each point is using either equation...

y = 2 + x

When x=-5, y=-3
and
when x=1 , y=3

the two points are (x,y) = (-5,-3) and (x,y)=(1,3)

Answer 8

@DanJS Thanks for the help!