Solve the Inequality for x (x+2)/(x-3)>=0

$\frac{x + 2}{x - 3} \geq 0$

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x + 2 = 0$

$x - 3 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Add $3$ to both sides of the equation.

$x = 3$

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = - 2$

$x = 3$

Consolidate the solutions.

$x = - 2, 3$

Find the domain of $\frac{x + 2}{x - 3}$ .

Set the denominator in $\frac{x + 2}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The domain is all values of $x$ that make the expression defined.

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

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 3$

$x > 3$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$\frac{(- 4) + 2}{(- 4) - 3} \geq 0$

The left side $0.\overline{285714}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\frac{(0) + 2}{(0) - 3} \geq 0$

The left side $- 0.\overline{6}$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{(6) + 2}{(6) - 3} \geq 0$

The left side $2.\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

$x < - 2$ True

$- 2 < x < 3$ False

$x > 3$ True

The solution consists of all of the true intervals.

$x \leq - 2$ or $x > 3$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 2 or x > 3$

Interval Notation:

$(- \infty, - 2] \cup (3, \infty)$

Solve the Inequality for x (x+2)/(x-3)>=0

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