Question

Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set.

$\frac{X-3}{X+1}\ge 0$

Short answer

The solution set is $(-\infty ,-1)\cup \left[3,\infty \right)$.

Step-by-step solution

Step 1. Finding the intervals.

Given inequality is

$\begin{array}{l}\frac{x-3}{x+1}\ge 0\\ \frac{(x-3)(x+1)}{{(x+1)}^2}\ge 0\\ (x-3)(x+1)\ge 0\end{array}$

Factors on the left-hand side of the inequality are $(x-3)(x+1)$.

These factors are zero when $x=3,-1$.

The factors divide the real line into 3 intervals$(-\infty ,-1),(-1,3)(3,\infty )$

Step 2. Determining the sign of each factor.

To determine the sign of each factor on each of the intervals that we found, we use test values. We choose a number inside each interval and check the sign of each factor.

Step 3. Making the table.

The value of inequality should be greater than or equal to zero.

Use open interval at$-1$ and closed interval at 3. Because at$-1$ denominator will be zero and the inequality will be not defined.

So, the solution set is$(-\infty ,-1)\cup \left[3,\infty \right)$

Step 4. Plotting the graph.

Graph of the inequality is: