Chapter EP, Problem 9.6.11EP

To determine

To solve: The given mathematical expression and check your solution.

Answer to Problem 9.6.11EP

  $z=-2$ is solution of the given mathematical expression.

Explanation of Solution

Given: $\frac{5z+2}{z^2-4}=\frac{-5z}{2-z}+\frac{2}{z+2}$

Explanation:Let $\frac{5z+2}{z^2-4}=\frac{-5z}{2-z}+\frac{2}{z+2}$ be the given mathematical expression and it has to solve.

  $\begin{array}{l}\frac{5z+2}{z^2-4}=\frac{-5z}{2-z}+\frac{2}{z+2}\\ \frac{5z+2}{z^2-4}=\frac{-5z}{-\left(z-2\right)}+\frac{2}{z+2}\\ \frac{5z+2}{z^2-4}=\frac{5z}{\left(z-2\right)}+\frac{2}{z+2}\\ \frac{5z+2}{z^2-4}=\frac{5z\left(z+2\right)+2\left(z-2\right)}{\left(z^2-4\right)}\end{array}$

  $\begin{array}{l}5z+2=5z^2+10z+2z-4\\ 5z+2=5z^2+12z-4\\ 5z^2+12z-4-5z-2=0\end{array}$

  $\begin{array}{l}5z^2+7z-6=0\\ 5z^2+10z-3z-6=0\\ 5z\left(z+2\right)-3\left(z+2\right)=0\\ \left(5z-3\right)\left(z+2\right)=0\end{array}$

  $\begin{array}{l}\left(5z-3\right)\left(z+2\right)=0\\ 5z-3=0\\ 5z=3\\ z=\frac{3}{5}\\ z+2=0\\ z=-2\end{array}$

Now it has to check the solution

For check solution put the answer $z=-2$ in equation

  $\begin{array}{l}\frac{5z+2}{z^2-4}=\frac{-5z}{2-z}+\frac{2}{z+2}\\ \frac{5\left(-2\right)+2}{{\left(-2\right)}^2-4}=\frac{-5\left(-2\right)}{2-\left(-2\right)}+\frac{2}{\left(-2\right)+2}\\ \frac{-10+2}{4-4}=\frac{10}{4}+\frac{2}{\left(-2\right)+2}\end{array}$

  $\begin{array}{l}\infty =\frac{10}{4}+\infty \\ \infty =\infty \end{array}$

Hence solution of the given mathematical expression is $z=-2$.