Question

Solve each inequality algebraically.

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

Short answer

Solution to the inequality$\frac{x-3}{x+1}>0$ is $(-\infty ,-1)\cup (3,\infty )$.

Step-by-step solution

Step 1. Given information

We have been given an inequality $\frac{x-3}{x+1}>0$.

We have to solve this inequality algebraically.

Step 2. Determine the real numbers at which the expression f equals zero and at which the expression f is undefined. 

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

$$\begin{gathered} x-3=0 \\ x=3 \\ \mathrm{Also},x+1=0 \\ x=-1 \end{gathered}$$

Step 3. Form the intervals 

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

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

Step 4. Select a number in each interval and evaluate f at the number 

Create the following table:

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

Step 5. Identify the interval

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