Question

Solve each inequality algebraically.

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

Short answer

Solution to the inequality $\frac{5}{x-3}>\frac{3}{x+1}$is $(-7,-1)\cup (3,\infty )$.

Step-by-step solution

Step 1. Given information  

We have been given an inequality $\frac{5}{x-3}>\frac{3}{x+1}$.

We have to solve this inequality algebraically.

Step 2. Write the inequality so that a polynomial or rational expression f is on the left side and zero is on the right side  

$$\begin{gathered} \frac{5}{x-3}>\frac{3}{x+1} \\ \frac{5}{x-3}-\frac{3}{x+1}>0 \\ \frac{5(x+1)-3(x-3)}{(x-3)(x+1)}>0(\mathrm{Take}\mathrm{LCD}) \\ \frac{5x+5-3x+9}{(x-3)(x+1)}>0 \\ \frac{2x+14}{(x-3)(x+1)}>0 \end{gathered}$$

Step 3. Determine the real numbers at which the expression f equals zero and at which the expression f is undefined. 

Assume $f\left(x\right)=\frac{2x+14}{(x-3)(x+1)}$.

$$\begin{gathered} 2x+14=0 \\ 2x=-14 \\ x=-7 \\ x-3=0 \\ x=3 \\ x+1=0 \\ x=-1 \end{gathered}$$

Step 4. Form the intervals  

Using the values of x found in previous step, we can divide the real numbers in the intervals:

$(-\infty ,-7)\cup (-7,-1)\cup (-1,3)\cup (3,\infty )$

Step 5. Select a number in each interval and evaluate f at the number   

Create the following table:

Interval	$(-\infty ,-7)$	$(-7,-1)$	$(-1,3)$	$(3,\infty )$
Number chosen	$-8$	$-2$	0	4
Value of f	$f(-8)=-\frac{1}{7}$	$f(-2)=1$	$f\left(0\right)=-\frac{7}{3}$	$f\left(4\right)=\frac{11}{5}$
Conclusion	negative	positive	negative	positive

Step 6. Identify the interval 

Since we want to know where f is positive, we conclude that $f\left(x\right)>0$ in the interval $(-7,-1)\cup (3,\infty )$.