Question

In the following exercise, simplify

$\frac{{\displaystyle \frac{2}{x^2-49}}+{\displaystyle \frac{1}{z+7}}}{{\displaystyle \frac{9}{z+7}}+{\displaystyle \frac{12}{z-7}}}$

Short answer

The solution is,$\frac{z-5}{21z+21}$

Step-by-step solution

Step 1. Given Information 

$\frac{{\displaystyle \frac{2}{x^2-49}+\frac{1}{z+7}}}{{\displaystyle \frac{9}{z+7}+\frac{12}{z-7}}}$

Step 2. Simplify.

The LCD of all fractions in the complex rational expression is $\left(z+7\right)\left(z-7\right)$

Multiply the numerator and denominator by the LCD.

$$\begin{gathered} =\frac{{\displaystyle \left(z+7\right)\left(z-7\right)\left(\frac{2}{x^2-49}+\frac{1}{z+7}\right)}}{{\displaystyle \left(z+7\right)\left(z-7\right)\left(\frac{9}{z+7}+\frac{12}{z-7}\right)}} \\ =\frac{{\displaystyle \left(\left(z+7\right)\left(z-7\right)\frac{2}{x^2-49}+\left(z+7\right)\left(z-7\right)\frac{1}{z+7}\right)}}{{\displaystyle \left(\left(z+7\right)\left(z-7\right)\frac{9}{z+7}+\left(z+7\right)\left(z-7\right)\frac{12}{z-7}\right)}} \\ =\frac{2+\left(z-7\right)}{9\left(z-7\right)+12\left(z+7\right)} \\ =\frac{z-5}{21z+21} \end{gathered}$$