Question

Subtract$\frac{v}{v-3}-\frac{v-6}{v+1}$.

Short answer

The difference is$\frac{2(5v-9)}{(v-3)(v+1)}$.

Step-by-step solution

Given Information  

The Given expression is$\frac{v}{v-3}-\frac{v-6}{v+1}$.

Find the LCD of denominators.  

Since there are no common denominators.

The LCD is$(v-3)(v+1)$.

Rewrite each fraction as an equivalent fraction with the LCD.  

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

Simplify.

$=\frac{v^2+v}{(v-3)(v+1)}-\frac{v^2-3v-6v+18}{(v-3)(v+1)}$

Subtract and simplify the numerator. 

Take common denominator.

$=\frac{v^2+v-(v^2-9v+18)}{(v-3)(v+1)}$

Simplify numerator.

$=\frac{v^2+v-v^2+9v-18}{(v-3)(v+1)}$

$=\frac{10v-18}{(v-3)(v+1)}$

$=\frac{2(5v-9)}{(v-3)(v+1)}$