Solve the Inequality for x (x-4)/(x+3)>0

$\frac{x - 4}{x + 3} > 0$

Step 1

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

$x - 4 = 0$

$x + 3 = 0$

Step 2

Add $4$ to both sides of the equation.

$x = 4$

Step 3

Subtract $3$ from 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.

$x = 4$

$x = - 3$

Step 5

Consolidate the solutions.

$x = 4, - 3$

Step 6

Find the domain of $\frac{x - 4}{x + 3}$ .

Set the denominator in $\frac{x - 4}{x + 3}$ equal to $0$ to find where the expression is undefined.

$x + 3 = 0$

Subtract $3$ from 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.

$x < - 3$

$- 3 < x < 4$

$x > 4$

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 $x < - 3$ 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.

$\frac{(- 6) - 4}{(- 6) + 3} > 0$

The left side $3.\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $- 1.\overline{3}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

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

$\frac{(6) - 4}{(6) + 3} > 0$

The left side $0.\overline{2}$ 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.

$x < - 3$ True

$- 3 < x < 4$ False

$x > 4$ True

$x < - 3$ True

$- 3 < x < 4$ False

$x > 4$ True

Step 9

The solution consists of all of the true intervals.

$x < - 3$ or $x > 4$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or x > 4$

Interval Notation:

$(- \infty, - 3) \cup (4, \infty)$

Step 11

Solve the Inequality for x (x-4)/(x+3)>0

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