# How do you simplify 2/(x-1) + (x-11)/(x^2+3x-4)?

Mar 18, 2016

$\frac{3}{x + 4}$

#### Explanation:

$1$. Factor the denominator of the second fraction.

$\frac{2}{x - 1} + \frac{x - 11}{{x}^{2} + 3 x - 4}$

$= \frac{2}{x - 1} + \frac{x - 11}{\left(x + 4\right) \left(x - 1\right)}$

$2$. To add the two fractions together, find its L.C.D. (lowest common denominator).

$= \frac{2 \textcolor{b l u e}{\left(x + 4\right)}}{\left(x - 1\right) \textcolor{b l u e}{\left(x + 4\right)}} + \frac{x - 11}{\left(x + 4\right) \left(x - 1\right)}$

$= \frac{2 \textcolor{b l u e}{\left(x + 4\right)} + \left(x - 11\right)}{\left(x + 4\right) \left(x - 1\right)}$

$3$. Simplify the numerator.

$= \frac{2 x + 8 + x - 11}{\left(x + 4\right) \left(x - 1\right)}$

$= \frac{3 x - 3}{\left(x + 4\right) \left(x - 1\right)}$

$4$. Factor out $3$ from the numerator.

$= \frac{3 \left(x - 1\right)}{\left(x + 4\right) \left(x - 1\right)}$

$5$. Since the factor, $\left(x - 1\right)$, appears in both the numerator and denominator, they cancel out.

$= \frac{3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}}}{\left(x + 4\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(x - 1\right)}}}}$

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \frac{3}{x + 4} \textcolor{w h i t e}{\frac{a}{a}} |}}}$