Question 486725
Your work so far is good!
1) {{{x^2 = x+y}}} and...
2) {{{y = 2x-2}}} Substitute this for y in equaton 1) to get...
1a) {{{x^2 = x+(2x-2)}}} Simplify and solve for x:
1b) {{{x^2-3x+2 = 0}}} Solve by factoring:
1c) {{{(x-1)(x-2) = 0}}} so...
1d) {{{highlight(x = 1)}}} or {{{highlight(x = 2)}}} Now plug each of these, in turn, in equation 2 and solve for y.
2a) {{{y = 2x-2}}} Substitute {{{x = 1}}}:
2b) {{{highlight(y = 0)}}} and...
2c) {{{y = 2x-2}}} Substitute {{{x = 2}}}:
2d) {{{highlight(y = 2)}}}
So there are four possible solutions:
(1, 0), (1, 2), (2, 0), (2, 2)
Let's check them all:
{{{log(x+y) = 2Log(x)}}} Substitute {{{x = 1}}} and {{{y = 0}}}
{{{Log(1) = 2Log(1)}}}
{{{0 = 0}}} This solution is valid!
Try the other solutions yourself and see what happens!