Chapter 9.6, Problem 27PPS

To determine

To calculate: The given inequality and check solution.

Answer to Problem 27PPS

the solution is $x<-2$ or $2<x<14$ .

Explanation of Solution

Given information: Inequality $\frac{x-2}{x+2}+\frac{1}{x-2}>\frac{x-4}{x-2}$ .

Calculation:

Inequality $\frac{x-2}{x+2}+\frac{1}{x-2}>\frac{x-4}{x-2}$ .

The excluded value for this inequality are $2$ and $-2$ .

Solve the related equation.

  $\frac{x-2}{x+2}+\frac{1}{x-2}=\frac{x-4}{x-2}$

The LCD for the terms is $\left(x+2\right)\times \left(x-2\right)$ .

Multiply both side by LCD

  $\begin{array}{l}\left(x-2\right)\left(x-2\right)+\left(x+2\right)=\left(x-4\right)\left(x+2\right)\\ x^2+4-4x+x+2=x^2-2x-8\\ -x=-14\\ x=14\end{array}$

Draw vertical lines at the excluded value and at the solutions to separates the number line into intervals.

  

Now test a sample value in each interval to determine whether the values in the interval satisfy the inequality.

Test $x=-3$

  $\begin{array}{l}\frac{-3-2}{-3+2}+\frac{1}{-3-2}>\frac{-3-4}{-3-2}\\ \frac{-5}{-1}+\frac{1}{-5}>\frac{-7}{-5}\\ \frac{5}{1}-\frac{1}{5}>\frac{7}{5}\\ \frac{24}{5}>\frac{7}{5}\end{array}$

Test $x=3$

  $\begin{array}{l}\frac{3-2}{3+2}+\frac{1}{3-2}>\frac{3-4}{3-2}\\ \frac{1}{5}+\frac{1}{1}>\frac{-1}{1}\\ \frac{6}{5}>-1\end{array}$

Test $x=15$

  $\begin{array}{l}\frac{15-2}{15+2}+\frac{1}{15-2}>\frac{15-4}{15-2}\\ \frac{13}{17}+\frac{1}{13}>\frac{11}{13}\\ \frac{186}{221}>\frac{11}{13}\end{array}$

The statement is true for $x=-3$ and $x=3$ .Therefore, the solution is $x<-2$ or $2<x<14$ .