Chapter 4, Problem 11CT

To determine

The solution of the inequality $\frac{x+2}{x-3}<2$ algebraically.

Answer to Problem 11CT

Solution:

The solution of the inequality $\frac{x+2}{x-3}<2$ is $\left(-\infty ,3\right)\cup \left(8,\infty \right)$ .

Explanation of Solution

Given information:

The inequality, $\frac{x+2}{x-3}<2$ .

Step 1: Write the inequality so that the polynomial expression $f$ on left side and zero is on the right side.

  $\frac{x+2}{x-3}<2$

Subtract $2$ on both sides.

  $\frac{x+2}{x-3}-2<0$

Multiply $2$ by $\frac{x-3}{x-3}$ ,

  $\frac{x+2}{x-3}-2\left(\frac{x-3}{x-3}\right)<0$

  $\frac{x+2-2x+6}{x-3}<0$

  $\frac{-x+8}{x-3}<0$

  $f(x)=\frac{-x+8}{x-3}$ .

Step 2: To determine the real zeros of the function and the real number where the function is undefined.

Finding the real zeros of $f$ .

  $\frac{-x+8}{x-3}=0$

  $-x+8=0$

  $x=8$ .

The zero of the function is $x=8$ .

The function $f(x)$ is undefined when $x-3=0$ that means at $x=3$ .

Step 3: Use the real zeroes and undefined value to separate the real number line into four intervals as $(-\infty ,3),(3,8),(8,\infty )$ .

Step 4: To select a test point from each interval and determine whether $f(x)$ is positive or negative.

  $\begin{array}{cccc}\text{Interval}& (-\infty ,3)& (3,8)& (8,\infty )\\ \text{Number}\text{chosen}& 0& 4& 9\\ \text{Value}\text{of}f& \begin{array}{l}f(0)=\frac{0+8}{0-3}\\ =-\frac{8}{3}\end{array}& \begin{array}{l}f(4)=\frac{-4+8}{4-3}\\ =4\end{array}& \begin{array}{l}f(0)=\frac{-9+8}{9-3}\\ =-\frac{1}{6}\end{array}\\ \text{Conclusion}& \text{Negative}& \text{Positive}& \text{Negative}\end{array}$

  $f(x)$ is negative for the interval $\left(-\infty ,3\right)\text{or}\left(8,\infty \right)$ .

The inequality $\frac{x+2}{x-3}<2$ has the solution $\left(-\infty ,3\right)\text{or}\left(8,\infty \right)$ .

Therefore, the solution set for the inequality is $\left(-\infty ,3\right)\cup \left(8,\infty \right)$ .