# How do you simplify (x^2-5)/(x^2 +5x - 14)-((x+3)/(x+7))?

Jul 25, 2015

Here's how you could simplify this expression.

#### Explanation:

Start by writing out your starting expression

$\frac{{x}^{2} - 5}{{x}^{2} + 5 x - 14} - \frac{x + 3}{x + 7}$

Next, factor the denominator of the first fraction

${x}^{2} + 5 x - 14$

${x}^{2} + 7 x - 2 x - 14$

$x \left(x - 2\right) + 7 \left(x - 2\right)$

$\left(x - 2\right) \left(x + 7\right)$

Your expression is thus equivalent to

$\frac{{x}^{2} - 5}{\left(x - 2\right) \cdot \left(x + 7\right)} - \frac{x + 3}{x + 7}$

Since you have to subtract two fractions, you need to find the commonon denominator first. To do that, multiply the second fraction by $\frac{x - 2}{x - 2}$

$\frac{{x}^{2} - 5}{\left(x - 2\right) \cdot \left(x + 7\right)} - \frac{\left(x + 3\right) \cdot \left(x - 2\right)}{\left(x - 2\right) \cdot \left(x + 7\right)}$

This will get you

$\frac{{x}^{2} - 5 - \left(x + 3\right) \left(x - 2\right)}{\left(x - 2\right) \left(x + 7\right)}$

$\frac{\cancel{{x}^{2}} - 5 - \cancel{{x}^{2}} - x + 6}{\left(x - 2\right) \left(x + 7\right)} = \textcolor{g r e e n}{\frac{1 - x}{\left(x - 2\right) \left(x + 7\right)}}$