Question

Simplify:$\frac{\mathrm{v}}{v+1}+\frac{3}{v-1}-\frac{6}{v^2-1}$

Short answer

The simplification of the rational expression is$\frac{{\mathrm{v}}^2+2\mathrm{v}-3}{\left(v+1\right)\left(v-1\right)}$.

Step-by-step solution

Given information

The given rational expression is$\frac{v}{v+1}+\frac{3}{v-1}-\frac{6}{v^2-1}$.

The simplification of the rational expression 

$\frac{v}{v+1}+\frac{3}{v-1}-\frac{6}{\left(v-1\right)\left(v+1\right)}$

The LCD of the expression is $\left(v+1\right)\left(v-1\right)$.

Rewrite each rational expression as an equivalent rational expression with the LCD.

$$\begin{gathered} =\frac{v\left(v-1\right)}{\left(v+1\right)\left(v-1\right)}+\frac{3\left(v+1\right)}{\left(v-1\right)\left(v+1\right)}-\frac{6}{\left(v+1\right)\left(v-1\right)} \\ =\frac{v^2-v}{\left(v+1\right)\left(v-1\right)}+\frac{3v+3}{\left(v+1\right)\left(v-1\right)}-\frac{6}{\left(v+1\right)\left(v-1\right)} \\ Simplify \\ =\frac{v^2-v+3v+3-6}{\left(v+1\right)\left(v-1\right)} \\ =\frac{v^2+2v-3}{\left(v+1\right)\left(v-1\right)} \\ \left({\mathrm{v}}^2+2\mathrm{v}-3\right)\mathrm{cannot}\mathrm{factorized}\mathrm{further}. \\ \frac{{\mathrm{v}}^2+2\mathrm{v}-3}{\left(\mathrm{v}+1\right)\left(\mathrm{v}-1\right)} \end{gathered}$$