Chapter 5.9, Problem 13WE

To determine

To solve: the given equation for $\mathrm{y}$ . Check the result.

Answer to Problem 13WE

No solution

Explanation of Solution

Given information:

Given equation is

  $\frac{1}{\mathrm{y}-2}+\frac{1}{\mathrm{y}+2}=\frac{4}{{\mathrm{y}}^2-4}$

Calculation:

Consider the given equation,

  $\frac{1}{\mathrm{y}-2}+\frac{1}{\mathrm{y}+2}=\frac{4}{{\mathrm{y}}^2-4}$

Now, solve the equation as shown,

  $\begin{array}{c}\frac{1}{\mathrm{y}-2}+\frac{1}{\mathrm{y}+2}=\frac{4}{{\mathrm{y}}^2-4}\\ \frac{\mathrm{y}+2+\mathrm{y}-2}{\left(\mathrm{y}+2\right)\left(\mathrm{y}-2\right)}=\frac{4}{{\mathrm{y}}^2-4}\\ \frac{2\mathrm{y}}{{\mathrm{y}}^2-4}=\frac{4}{{\mathrm{y}}^2-4}\\ 4\left({\mathrm{y}}^2-4\right)=2\mathrm{y}\left({\mathrm{y}}^2-4\right)\\ \left({\mathrm{y}}^2-4\right)\left(2\mathrm{y}-4\right)=0\\ \mathrm{y}=2,-2,2\end{array}$

Now, check the result by back substitution.

  $\begin{array}{c}\frac{1}{\mathrm{y}-2}+\frac{1}{\mathrm{y}+2}=\frac{4}{{\mathrm{y}}^2-4}\\ \frac{1}{2-2}+\frac{1}{2+2}=\frac{4}{2^2-4}\\ \frac{1}{0}+\frac{1}{4}=\frac{4}{0}\end{array}$

Which is undefined.

Hence, the equation has no solution.