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

Jun 10, 2015

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

Explanation:

Have a look:

Jun 10, 2015

Multiply through by $\left(x + 2\right)$ and $\left(x - 2\right)$, expand and simplify to get:
$0 = {x}^{2} - 2 x - 24 = \left(x - 6\right) \left(x + 4\right)$, hence $x = 6$ or $x = - 4$

Explanation:

Given:

$\frac{2 x}{x + 2} - 2 = \frac{x - 8}{x} - 2$

Multiply both sides by $\left(x + 2\right) \left(x - 2\right)$ to get:

$2 x \left(x - 2\right) - 2 \left(x + 2\right) \left(x - 2\right) = \left(x - 8\right) \left(x + 2\right)$

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

Multiplying out:

$\cancel{2 {x}^{2}} - 4 x - \cancel{2 {x}^{2}} + 8 = {x}^{2} - 6 x - 16$

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

$0 = {x}^{2} - 2 x - 24 = \left(x - 6\right) \left(x + 4\right)$

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

$\left(x - a\right) \left(x + b\right) = {x}^{2} - \left(a - b\right) x - \left(a \times b\right)$

Since $\left(x - 6\right) \left(x + 4\right) = 0$, the solutions are $x = - 4$ and $x = 6$.