Chapter 9, Problem 26MS

To determine

To Simplify: The expression and assume that no denominator is zero.

Answer to Problem 26MS

  $\frac{4(v+3)}{(v+2)(v+4)}$

Explanation of Solution

Given: $\frac{v+4}{v+2}-\frac{v+2}{v+4}$

Given: $\frac{v+4}{v+2}-\frac{v+2}{v+4}$

Solving by taking LCM

  $\begin{array}{l}=\frac{v+4}{v+2}-\frac{v+2}{v+4}\\ =\frac{(v+4)(v+4)-(v+2)(v+2)}{(v+2)(v+4)}\\ =\frac{v^2+8v+16-v^2-4v-4}{(v+2)(v+4)}\\ =\frac{4v+12}{(v+2)(v+4)}\\ =\frac{4(v+3)}{(v+2)(v+4)}\end{array}$

Hence, $\frac{4(v+3)}{(v+2)(v+4)}$