Chapter SH, Problem 11.8.18EP

To determine

The solution to the given equation.

Answer to Problem 11.8.18EP

  $z=3$

Explanation of Solution

Given:

The given equation is:

  $\frac{z+3}{z-1}+\frac{z+1}{z-3}=2$

Calculation:

The solution for the given equation can be obtained as:

  $\begin{array}{l}\frac{z+3}{z-1}+\frac{z+1}{z-3}=2\\ \frac{(z+3)(z-3)+(z+1)(z-1)}{(z-1)(z-3)}=2\\ \frac{2z^2-10}{z^2-4z+3}=2\\ 2z^2-10=2z^2-8z-6\\ 8z=16\\ z=2\end{array}$

Check the solution in the original equation.

Let $z=1$ ,

  $\begin{array}{l}\frac{z+3}{z-1}+\frac{z+1}{z-3}=2\\ \frac{1+3}{1-1}+\frac{1+1}{1-3}=2\\ \frac{4}{0}-1\ne 2\\ \end{array}$

Hence, 1 is an extraneous solution.