# How do you simplify (x-1)/(x^2-1)+2/(5x+5)?

Jul 22, 2016

$\frac{7}{5 \left(x + 1\right)}$

#### Explanation:

$\frac{x - 1}{{x}^{2} - 1} + \frac{2}{5 x + 5}$

Focus on the first fraction first. Notice that the denominator is the difference of two squares, since

${x}^{2} - 1 = {x}^{2} - {1}^{2}$

As you know, the difference of two squares can be factored as

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that the first fraction is equivalent to

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}} \left(x + 1\right)} = \frac{1}{x + 1}$

Next, focus on the second fraction. In this case, the denominator can be factored as

$\frac{2}{5 x + 5} = \frac{2}{5 \left(x + 1\right)}$

The expression can thus be written as

$\frac{1}{x + 1} + \frac{2}{5 \left(x + 1\right)}$

The next thing to do here is multiply the first fraction by $1 = \frac{5}{5}$. This will allow you to express both fractions in terms of their common denominator, which is equal to $5 \left(x + 1\right)$

$\frac{1}{x + 1} \cdot \frac{5}{5} + \frac{2}{5 \left(x + 1\right)} = \frac{5}{5 \left(x + 1\right)} + \frac{2}{5 \left(x + 1\right)}$

$= \frac{2 + 5}{5 \left(x + 1\right)} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{7}{5 \left(x + 1\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Notice that you must have $x \ne \pm 1$.