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

May 31, 2017

$x = 0 , - 1$

#### Explanation:

Solve:

$\frac{{x}^{2} - 3}{\textcolor{b l u e}{\left(x + 2\right)}} = \frac{x - 3}{\textcolor{red}{2}}$

Cross multiply.

$\textcolor{red}{2} \left({x}^{2} - 3\right) = \textcolor{b l u e}{\left(x + 2\right)} \left(x - 3\right)$

Expand.

$2 {x}^{2} - 6 = \left(x + 2\right) \left(x - 3\right)$

FOIL the right side.

$2 {x}^{2} - 6 = {x}^{2} - 3 x + 2 x - 6$

Simplify.

$2 {x}^{2} - 6 = {x}^{2} - x - 6$

Add $6$ to both sides.

$2 {x}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} = {x}^{2} - x - \textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{6}}}$

Simplify.

$2 {x}^{2} = {x}^{2} - x$

Move all terms to the left side.

$2 {x}^{2} - {x}^{2} + x = 0$

Simplify.

${x}^{2} + x = 0$

Factor out the $x$.

$x \left(x + 1\right) = 0$

Solutions for $x$.

$x = 0$

$x = - 1$