# How do you solve (2x-3) /( x-1) - (x-1 )/( x+2) = (2x-5) /( x+2) + (2-x) /( 1-x) ?

Oct 18, 2015

$x = \frac{8}{5}$

#### Explanation:

Mutiply both sides by $\left(x - 1\right) \left(x + 2\right)$

$\left(2 x - 3\right) \left(x + 2\right) - {\left(x - 1\right)}^{2} = \left(2 x - 5\right) \left(x - 1\right) + \left(x - 2\right) \left(x + 2\right)$

Simplify

${x}^{2} + 3 x - 7 = 3 {x}^{2} - 7 x + 1$
$2 {x}^{2} - 10 x + 8 = 0$
${x}^{2} - 5 x + 4 = 0$
$\left(x - 1\right) \left(x - 4\right) = 0$
$x = 1 \mathmr{and} x = 4$

$x$ must differ from $1$ or we'd have a division by zero, so $x = 0$