Please explain ---The straight line L has equation y=3x-1 and the curve C has equation y=(x+3)(x-1)

The straight line L has equation y=3x-1 and the curve C has equationy=(x+3)(x-1). Show that the x-coordinates of the points of intersection of L and C satisfy the equation x^2-x-2=0

How do I work this out?

thank you

given.
y = 3x -1 ...............(1)
and y = (x +3)(x -1)...........(2)
then (x+3)(x-1) = 3x -1
x^2 -x +3x -3 = 3x - 1
x^2 - x +3x -3 -3x + 1 = 0
ie x^2 -x -2 = 0 - this is the required result
factorising gives ( x -2)(x + 1) = 0 =======> x = 2 and x = -1
from ( 1) when x = 2 , y = 5 so ( 2,5) is a co-ordinate of one of the intersections
when x = -1 . y = -4 so ( -1, -4) is the co-ordinates of the other point of intersection

Your best bet is to graph both equations. A graphing calculator works well but both equations will fit on a grid of x=0 to 10 and y= 0 to 10. Where the equations intersect take the x value and input it into x^2-x-2=0. Incidentally the intersection is at (2,5), so input 2 for x -> 2^2-2-2=0 or 4-4=0. It all works out.

3x - 1 = (x + 3)(x - 1)
3x - 1 = x² + 2x - 3
x² - x - 2 = 0