Solve for x (2x-3)/(3x+2)<0

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

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

$2 x - 3 = 0$

$3 x + 2 = 0$

Add $3$ to both sides of the equation.

$2 x = 3$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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.

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

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

Consolidate the solutions.

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

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

Set the denominator in $\frac{2 x - 3}{3 x + 2}$ 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.

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

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

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

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 $x < - \frac{2}{3}$ 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.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

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

The left side $1.\overline{285714}$ is not less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

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

The left side $- 1.5$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 4$

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

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

The left side $0.3\overline{571428}$ is not less than the right side $0$ , which means that the given statement is false.

False

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

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

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

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

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

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

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

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Solve for x (2x-3)/(3x+2)<0

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