Question

How do you solve `(2x)/ (x+2) - 2 = (x-8) / (x-2)`?

Answer 1

I would find a common denominator and then solve for `x`.

Explanation:

Have a look:

Answer 2

Multiply through by `(x+2)` and `(x-2)`, expand and simplify to get:
`0 = x^2-2x-24 = (x-6)(x+4)`, hence `x = 6` or `x=-4`

Explanation:

Given:

`(2x)/(x+2)-2 = (x-8)/x-2`

Multiply both sides by `(x+2)(x-2)` to get:

`2x(x-2)-2(x+2)(x-2) = (x-8)(x+2)`

Note that when we multiply by `(x+2)(x-2)` we could in theory introduce spurious extra solutions `x=+-2`, but these are excluded values of the original equation anyway.

Multiplying out:

`cancel(2x^2)-4x-cancel(2x^2)+8 = x^2-6x-16`

Add `4x-8` to both sides to get:

`0 = x^2-2x-24 = (x-6)(x+4)`

To get this factorization, I looked for a pair of factors of `24` whose difference is `-2`, since:

`(x - a)(x + b) = x^2-(a-b)x-(axxb)`

Since `(x-6)(x+4) = 0`, the solutions are `x=-4` and `x=6`.