Solve the Inequality for x (x-9)/(3x+2)>=3

$\frac{x - 9}{3 x + 2} \geq 3$

Subtract $3$ from both sides of the inequality.

$\frac{x - 9}{3 x + 2} - 3 \geq 0$

Simplify $\frac{x - 9}{3 x + 2} - 3$ .

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{3 x + 2}{3 x + 2}$ .

$\frac{x - 9}{3 x + 2} - 3 \cdot \frac{3 x + 2}{3 x + 2} \geq 0$

Combine $- 3$ and $\frac{3 x + 2}{3 x + 2}$ .

$\frac{x - 9}{3 x + 2} + \frac{- 3 (3 x + 2)}{3 x + 2} \geq 0$

Combine the numerators over the common denominator.

$\frac{x - 9 - 3 (3 x + 2)}{3 x + 2} \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{x - 9 - 3 (3 x) - 3 \cdot 2}{3 x + 2} \geq 0$

Multiply $3$ by $- 3$ .

$\frac{x - 9 - 9 x - 3 \cdot 2}{3 x + 2} \geq 0$

Multiply $- 3$ by $2$ .

$\frac{x - 9 - 9 x - 6}{3 x + 2} \geq 0$

Subtract $9 x$ from $x$ .

$\frac{- 8 x - 9 - 6}{3 x + 2} \geq 0$

Subtract $6$ from $- 9$ .

$\frac{- 8 x - 15}{3 x + 2} \geq 0$

Factor $- 1$ out of $- 8 x$ .

$\frac{- (8 x) - 15}{3 x + 2} \geq 0$

Rewrite $- 15$ as $- 1 (15)$ .

$\frac{- (8 x) - 1 (15)}{3 x + 2} \geq 0$

Factor $- 1$ out of $- (8 x) - 1 (15)$ .

$\frac{- (8 x + 15)}{3 x + 2} \geq 0$

Rewrite $- (8 x + 15)$ as $- 1 (8 x + 15)$ .

$\frac{- 1 (8 x + 15)}{3 x + 2} \geq 0$

Move the negative in front of the fraction.

$- \frac{8 x + 15}{3 x + 2} \geq 0$

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

$8 x + 15 = 0$

$3 x + 2 = 0$

Subtract $15$ from both sides of the equation.

$8 x = - 15$

Divide each term in $8 x = - 15$ by $8$ and simplify.

Divide each term in $8 x = - 15$ by $8$ .

$\frac{8 x}{8} = \frac{- 15}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{- 15}{8}$

Divide $x$ by $1$ .

$x = \frac{- 15}{8}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{15}{8}$

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Divide each term in $3 x = - 2$ by $3$ and simplify.

Divide each term in $3 x = - 2$ by $3$ .

$\frac{3 x}{3} = \frac{- 2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 2}{3}$

Divide $x$ by $1$ .

$x = \frac{- 2}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

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

$x = - \frac{15}{8}$

$x = - \frac{2}{3}$

Consolidate the solutions.

$x = - \frac{15}{8}, - \frac{2}{3}$

Find the domain of $- \frac{8 x + 15}{3 x + 2}$ .

Set the denominator in $\frac{8 x + 15}{3 x + 2}$ equal to $0$ to find where the expression is undefined.

$3 x + 2 = 0$

Solve for $x$ .

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Divide each term in $3 x = - 2$ by $3$ and simplify.

Divide each term in $3 x = - 2$ by $3$ .

$\frac{3 x}{3} = \frac{- 2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 2}{3}$

Divide $x$ by $1$ .

$x = \frac{- 2}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

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

$(- \infty, - \frac{2}{3}) \cup (- \frac{2}{3}, \infty)$

Use each root to create test intervals.

$x < - \frac{15}{8}$

$- \frac{15}{8} < x < - \frac{2}{3}$

$x > - \frac{2}{3}$

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 $x < - \frac{15}{8}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{15}{8}$ and see if this value makes the original inequality true.

$x = - 4$

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

$\frac{(- 4) - 9}{3 (- 4) + 2} \geq 3$

The left side $1.3$ is less than the right side $3$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{15}{8} < x < - \frac{2}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{15}{8} < x < - \frac{2}{3}$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

$\frac{(- 1) - 9}{3 (- 1) + 2} \geq 3$

The left side $10$ is greater than the right side $3$ , which means that the given statement is always true.

True

Test a value on the interval $x > - \frac{2}{3}$ to see if it makes the inequality true.

Choose a value on the interval $x > - \frac{2}{3}$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{(0) - 9}{3 (0) + 2} \geq 3$

The left side $- 4.5$ is less than the right side $3$ , which means that the given statement is false.

False

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

$x < - \frac{15}{8}$ False

$- \frac{15}{8} < x < - \frac{2}{3}$ True

$x > - \frac{2}{3}$ False

$x < - \frac{15}{8}$ False

$- \frac{15}{8} < x < - \frac{2}{3}$ True

$x > - \frac{2}{3}$ False

The solution consists of all of the true intervals.

$- \frac{15}{8} \leq x < - \frac{2}{3}$

The result can be shown in multiple forms.

Inequality Form:

$- \frac{15}{8} \leq x < - \frac{2}{3}$

Interval Notation:

$[- \frac{15}{8}, - \frac{2}{3})$

Solve the Inequality for x (x-9)/(3x+2)>=3

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