Question

What are the real and extraneous solutions in these 2 equations:
1. (x+5)/(x-3)=x

2. (2x)/(x^2)=(x-6)/(-4)

Answer 1

1. (x+5) = x^2 - 3x
x^2 - 4x - 5 = 0
(x-5)(x+1) so x = 5 and x = -1
neither are extraneous (both can substitute into the original)

Answer 2

wait what about number 2?

Answer 3

If you cross multiply -8x = x^3 - 6x^2
0 = x^3 - 6x^2 + 8x
x = 0
x^2 - 6x + 8 = 0
(x -4)(x - 2) = 0
x = 0, 2, 4
but x = 0 is extraneous (you cannot substitute it into the original problem without having an issue 0/0).