Solve the Inequality for x (3x+6)/(2x-12)>=0

$\frac{3 x + 6}{2 x - 12} \geq 0$

Step 1

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

$3 x + 6 = 0$

$2 x - 12 = 0$

Step 2

Subtract $6$ from both sides of the equation.

$3 x = - 6$

Step 3

Divide each term in $3 x = - 6$ by $3$ and simplify.

Divide each term in $3 x = - 6$ by $3$ .

$\frac{3 x}{3} = \frac{- 6}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 6}{3}$

Divide $x$ by $1$ .

$x = \frac{- 6}{3}$

Simplify the right side.

Divide $- 6$ by $3$ .

$x = - 2$

Step 4

Add $12$ to both sides of the equation.

$2 x = 12$

Step 5

Divide each term in $2 x = 12$ by $2$ and simplify.

Divide each term in $2 x = 12$ by $2$ .

$\frac{2 x}{2} = \frac{12}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{12}{2}$

Divide $x$ by $1$ .

$x = \frac{12}{2}$

Simplify the right side.

Divide $12$ by $2$ .

$x = 6$

Step 6

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

$x = - 2$

$x = 6$

Step 7

Consolidate the solutions.

$x = - 2, 6$

Step 8

Find the domain of $\frac{3 x + 6}{2 x - 12}$ .

Set the denominator in $\frac{3 x + 6}{2 x - 12}$ equal to $0$ to find where the expression is undefined.

$2 x - 12 = 0$

Solve for $x$ .

Add $12$ to both sides of the equation.

$2 x = 12$

Divide each term in $2 x = 12$ by $2$ and simplify.

Divide each term in $2 x = 12$ by $2$ .

$\frac{2 x}{2} = \frac{12}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{12}{2}$

Divide $x$ by $1$ .

$x = \frac{12}{2}$

Simplify the right side.

Divide $12$ by $2$ .

$x = 6$

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

$(- \infty, 6) \cup (6, \infty)$

Step 9

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 6$

$x > 6$

Step 10

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.

$\frac{3 (- 4) + 6}{2 (- 4) - 12} \geq 0$

The left side $0.3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 6$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{3 (0) + 6}{2 (0) - 12} \geq 0$

The left side $- 0.5$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$\frac{3 (8) + 6}{2 (8) - 12} \geq 0$

The left side $7.5$ 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

$- 2 < x < 6$ False

$x > 6$ True

$x < - 2$ True

$- 2 < x < 6$ False

$x > 6$ True

Step 11

The solution consists of all of the true intervals.

$x \leq - 2$ or $x > 6$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 2 or x > 6$

Interval Notation:

$(- \infty, - 2] \cup (6, \infty)$

Step 13

Solve the Inequality for x (3x+6)/(2x-12)>=0

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