Chapter 5.9, Problem 20WE

To determine

To solve and check , the given equation has no solution of $\frac{1}{s+3}+\frac{1}{s-5}=\frac{1-s}{s+3}$

Answer to Problem 20WE

The answer is $\left\{3,1\right\}$

Explanation of Solution

Given that: : $\frac{1}{s+3}+\frac{1}{s-5}=\frac{1-s}{s+3}$

Multiply by $(s+3)(s-5)$ , LCM of $(s+3)and(s-5)$

  $(s+3)(s-5)\left(\frac{1}{s+3}+\frac{1}{s-5}\right)=\left(\frac{1-s}{s+3}\right)(s-3)(s+5)$

Apply distributive property

  $\begin{array}{c}(s+3)(s-5)\left(\frac{1}{s+3}+\frac{1}{s-5}\right)=\left(\frac{1-s}{s+3}\right)(s+3)(s-5)\\ (s-5)+(s+3)=(1-s)(s-5)\\ s-5+s+3=(s-5-s^2+5s)\\ 2s-2=(-s^2+6s-5)\end{array}$

Add $(-2s+2)$ to both sides

  $\begin{array}{c}2s-2-2s+2=-s^2+6s-5-2s+2\\ 0=-s^2+4s-3\\ s^2-4s+3=0\end{array}$

Apply the factor method

  $\begin{array}{c}{s}^2-4s+3=0\\ s^2-(3+1)s+3=0\\ s^2-3s-s+3=0\\ s(s-3)-1(s-3)=0\\ (s-3)(s-1)=0\end{array}$

Apply zero product property to find the solution of equation

  $\begin{array}{c}(s-3)=0\\ s=3\\ \text{and}\text{similarly}\\ (s-1)=0\\ s=1\end{array}$

When s equal to -3 and 5, a denominator in the original equation has a value of 0. Therefore,

-3 and 5 are not in the domain of s and cannot be solution.

Check:

  $\begin{array}{c}Fors=3\\ \frac{1}{3+3}+\frac{1}{3-5}=\frac{1-3}{3+3}\\ \frac{1}{6}+\frac{1}{-2}=\frac{-2}{6}\\ \frac{-2}{6}=\frac{-2}{6}\\ \text{LHS=RHS}\end{array}$

  $\begin{array}{c}Fors=1\\ \frac{1}{1+3}+\frac{1}{1-5}=\frac{1-1}{1+3}\\ \frac{1}{4}+\frac{1}{-4}=\frac{0}{4}\\ 0=0\\ \text{LHS=RHS}\end{array}$

Hence, the solution of equation is $\left\{3,1\right\}$