Question

In the following exercises, solve.

$\frac{v-10}{v^2-5v+4}=\frac{3}{v-1}-\frac{6}{v-4}$

Short answer

The answer is$v=1.$

Step-by-step solution

Step 1. Given Information 

The given rational expression is

$\frac{v-10}{v^2-5v+4}=\frac{3}{v-1}-\frac{6}{v-4}$

Step 2. Eliminating the fractions 

The Least Common denominator is $(v-1)(v-4)$

Multiply the equation by LCD,

$$\begin{gathered} (v-1)(v-4)\left[\frac{v-10}{v^2-5v+4}\right]=(v-1)(v-4)[\frac{3}{v-1}-\frac{6}{v-4}] \\ {v}^2-5v+4\left[\frac{v-10}{v^2-5v+4}\right]=3(v-4)-6(v-1) \\ v-10=3v-12-6v+6 \\ v-10=-3v-6 \\ 4v-4=0 \\ 4v=4 \\ v=1 \end{gathered}$$