Chapter 9.6, Problem 30PPS

To determine

To calculate: The given inequality and check solution.

Answer to Problem 30PPS

the solution is $x<-5$ or $-2<x<1$ or $x>2$ .

Explanation of Solution

Given information: Inequality $\frac{x}{x+2}+\frac{1}{x-1}<\frac{3}{2}$ .

Calculation:

Inequality $\frac{x}{x+2}+\frac{1}{x-1}<\frac{3}{2}$ .

The excluded value for this inequality is $1$ and $-2$ .

Solve the related equation.

  $\frac{x}{x+2}+\frac{1}{x-1}=\frac{3}{2}$

The LCD for the terms is $2\times \left(x+2\right)\times \left(x-1\right)$ .

Multiply both side by LCD

  $\begin{array}{l}2x\times \left(x-1\right)+2\times \left(x+2\right)=3\left(x+2\right)\times \left(x-1\right)\\ -x^2-3x+10=0\\ x^2+3x-10=0\\ x=2,-5\end{array}$

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 $x=-6$

  $\begin{array}{l}\frac{-6}{-4}+\frac{1}{-6-1}<\frac{3}{2}\\ \frac{6}{4}-\frac{1}{6}<\frac{3}{2}\\ \frac{32}{24}<\frac{3}{2}\end{array}$

Test $x=-1$

  $\begin{array}{l}\frac{-1}{1}+\frac{1}{-1-1}<\frac{3}{2}\\ \frac{-1}{1}-\frac{1}{2}<\frac{3}{2}\\ \frac{-3}{2}<\frac{3}{2}\end{array}$

Test $x=\frac{3}{2}$

  $\begin{array}{l}\frac{\frac{3}{2}}{\frac{3}{2}+2}+\frac{1}{\frac{3}{2}-1}<\frac{3}{2}\\ \frac{3}{5}+2<\frac{3}{2}\\ \frac{13}{5}<\frac{3}{2}\times \end{array}$

Test $x=3$

  $\begin{array}{l}\frac{3}{3+2}+\frac{1}{3-1}<\frac{3}{2}\\ \frac{3}{5}+\frac{1}{2}<\frac{3}{2}\\ \frac{11}{10}<\frac{3}{2}\end{array}$

The statement is true for $x=-6$ , $x=-1$ and $x=3$ .Therefore, the solution is $x<-5$ or $-2<x<1$ or $x>2$ .