Solve the Inequality for x (x+1)/(x-3)>=0

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

Step 1

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

$x + 1 = 0$

$x - 3 = 0$

Step 2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3

Add $3$ to both sides of the equation.

$x = 3$

Step 4

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

$x = - 1$

$x = 3$

Step 5

Consolidate the solutions.

$x = - 1, 3$

Step 6

Find the domain of $\frac{x + 1}{x - 3}$ .

Set the denominator in $\frac{x + 1}{x - 3}$ 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.

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

Step 7

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 3$

$x > 3$

Step 8

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 $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

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

$\frac{(- 4) + 1}{(- 4) - 3} \geq 0$

The left side $0.\overline{428571}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{(0) + 1}{(0) - 3} \geq 0$

The left side $- 0.\overline{3}$ 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.

$\frac{(6) + 1}{(6) - 3} \geq 0$

The left side $2.\overline{3}$ 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.

$x < - 1$ True

$- 1 < x < 3$ False

$x > 3$ True

$x < - 1$ True

$- 1 < x < 3$ False

$x > 3$ True

Step 9

The solution consists of all of the true intervals.

$x \leq - 1$ or $x > 3$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 1 or x > 3$

Interval Notation:

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

Step 11

Solve the Inequality for x (x+1)/(x-3)>=0

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