Chapter 4.6, Problem 44AYU

In Problems 19-48, solve each inequality algebraically.

$\frac{5}{x\text{ - 3}}\text{ > }\frac{3}{x\text{ + 1}}$

To determine

To solve: The inequality algebraically.

Answer to Problem 44AYU

$\left\{x|-\infty <x<-7,-1<x<\text{ 3}\right\}$

Explanation of Solution

Given:

$\frac{5}{x-\text{ 3}}>\frac{3}{x+\text{ 1}}$

$\frac{5}{x-\text{ 3}}>\frac{3}{x+\text{ 1}}$

$\frac{5}{x-\text{ 3}}-\frac{3}{x+\text{ 1}}>\text{ 0}$

$\frac{5x+5-3x+9}{\left(x-3\right)\left(x+1\right)}>0$

$\frac{2x+14}{\left(x-3\right)\left(x+1\right)}>0$

The zero of $f\left(x\right)=\frac{2x+14}{\left(x-3\right)\left(x+1\right)}$ is $x=\text{ <mo>−</mo>7}$. Also, $f$ is undefined for $x=\text{ <mo>−</mo>1, 3}$.

Use the zero and the undefined value to separate the real number line into intervals:

$\left(-\infty ,-7\right),\left(-7,-1\right),\left(-1,\text{ 3}\right),\left(3,\infty \right)$

Select a test number in each interval found in above and evaluate $f\left(x\right)=\frac{2x+14}{\left(x-3\right)\left(x+1\right)}$ each number to determine if $f\left(x\right)$ is positive or negative.

Interval	$\left(-\infty ,\text{ <mo>−</mo>7}\right)$	$\left(-7,-1\right)$	$\left(-1,3\right)$	$\left(3,\infty \right)$
Number chosen	$-8$	$-2$	$2$	$4$
Value of $f$	$-\frac{2}{77}$	$2$	$-6$	$\frac{22}{5}$
Conclusion	Negative	Positive	Negative	Positive

Since we want to know where $f\left(x\right)$ is positive, conclude that $f\left(x\right)>0$. The solution set is $\left\{x|-\infty <x<-7,-1<x<\text{ 3}\right\}$ or, using interval notation $\left(-\infty ,-7\right)\cup \left(-1,\text{ 3}\right)$.