Chapter SH, Problem 11.8.10EP

To determine

The solution to the given equation.

Answer to Problem 11.8.10EP

  $x=\frac{1}{4}=0.25$

Explanation of Solution

Given:

The given equation is:

  $\frac{5x}{x+1}+\frac{1}{x}=5$

Calculation:

The solution for the given equation can be obtained as:

  $\begin{array}{l}\frac{5x}{x+1}+\frac{1}{x}=5\\ \frac{5x^2+x+1}{x^2+x}=5\\ 5x^2+x+1=5(x^2+x)\\ 5x^2+x+1=5x^2+5x\\ 5x-x=1\\ x=\frac{1}{4}=0.25\end{array}$

Check the solution in the original equation.

Let $x=0.25$ ,

  $\begin{array}{l}\frac{5x}{x+1}+\frac{1}{x}=5\\ \frac{5(0.25)}{0.25+1}+\frac{1}{0.25}=5\\ \frac{1.25}{1.25}+\frac{1}{0.25}=5\\ \frac{0.25+1}{0.25}=5\\ 5=5\end{array}$

So, $x=0.25$ is a solution.

Let $x=1$

  $\begin{array}{l}\frac{5x}{x+1}+\frac{1}{x}=5\\ \frac{5(1)}{1+1}+\frac{1}{1}=5\\ \frac{5}{2}+1=5\\ \frac{5+2}{2}=5\\ \frac{7}{2}\ne 5\end{array}$

Hence, 1 is an extraneous solution.