Finite Math Examples

$| \frac{x - 1}{2 x + 3} | < 1$

Step 1

Write $| \frac{x - 1}{2 x + 3} | < 1$ as a piecewise.

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Step 1.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$\frac{x - 1}{2 x + 3} \geq 0$

Step 1.2

Solve the inequality.

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Step 1.2.1

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

$x - 1 = 0$

$2 x + 3 = 0$

Step 1.2.2

Add $1$ to both sides of the equation.

$x = 1$

Step 1.2.3

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 1.2.4

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

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Step 1.2.4.1

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

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

Step 1.2.4.2

Simplify the left side.

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Step 1.2.4.2.1

Cancel the common factor of $2$ .

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Step 1.2.4.2.1.1

Cancel the common factor.

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

Step 1.2.4.2.1.2

Divide $x$ by $1$ .

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

Step 1.2.4.3

Simplify the right side.

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Step 1.2.4.3.1

Move the negative in front of the fraction.

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

Step 1.2.5

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

$x = 1$

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

Step 1.2.6

Consolidate the solutions.

$x = 1, - \frac{3}{2}$

Step 1.2.7

Find the domain of $| \frac{x - 1}{2 x + 3} | - 1$ .

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Step 1.2.7.1

Set the denominator in $\frac{x - 1}{2 x + 3}$ equal to $0$ to find where the expression is undefined.

$2 x + 3 = 0$

Step 1.2.7.2

Solve for $x$ .

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Step 1.2.7.2.1

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 1.2.7.2.2

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

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Step 1.2.7.2.2.1

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

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

Step 1.2.7.2.2.2

Simplify the left side.

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Step 1.2.7.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.7.2.2.2.1.1

Cancel the common factor.

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

Step 1.2.7.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.2.7.2.2.3

Simplify the right side.

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Step 1.2.7.2.2.3.1

Move the negative in front of the fraction.

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

Step 1.2.7.3

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

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

Step 1.2.8

Use each root to create test intervals.

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

$- \frac{3}{2} < x < 1$

$x > 1$

Step 1.2.9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.2.9.1

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

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Step 1.2.9.1.1

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

$x = - 4$

Step 1.2.9.1.2

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

$\frac{(- 4) - 1}{2 (- 4) + 3} \geq 0$

Step 1.2.9.1.3

The left side $1$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.9.2

Test a value on the interval $- \frac{3}{2} < x < 1$ to see if it makes the inequality true.

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Step 1.2.9.2.1

Choose a value on the interval $- \frac{3}{2} < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 1.2.9.2.2

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

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

Step 1.2.9.2.3

The left side $- 0.\overline{3}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.9.3

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

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Step 1.2.9.3.1

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

$x = 4$

Step 1.2.9.3.2

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

$\frac{(4) - 1}{2 (4) + 3} \geq 0$

Step 1.2.9.3.3

The left side $0.\overline{27}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.9.4

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

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

$- \frac{3}{2} < x < 1$ False

$x > 1$ True

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

$- \frac{3}{2} < x < 1$ False

$x > 1$ True

Step 1.2.10

The solution consists of all of the true intervals.

$x < - \frac{3}{2}$ or $x \geq 1$

Step 1.3

In the piece where $\frac{x - 1}{2 x + 3}$ is non-negative, remove the absolute value.

$\frac{x - 1}{2 x + 3} < 1$

Step 1.4

Find the domain of $\frac{x - 1}{2 x + 3} < 1$ and find the intersection with $x < - \frac{3}{2} or x \geq 1$ .

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Step 1.4.1

Find the domain of $\frac{x - 1}{2 x + 3} < 1$ .

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Step 1.4.1.1

Set the denominator in $\frac{x - 1}{2 x + 3}$ equal to $0$ to find where the expression is undefined.

$2 x + 3 = 0$

Step 1.4.1.2

Solve for $x$ .

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Step 1.4.1.2.1

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 1.4.1.2.2

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

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Step 1.4.1.2.2.1

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

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

Step 1.4.1.2.2.2

Simplify the left side.

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Step 1.4.1.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.4.1.2.2.2.1.1

Cancel the common factor.

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

Step 1.4.1.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.4.1.2.2.3

Simplify the right side.

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Step 1.4.1.2.2.3.1

Move the negative in front of the fraction.

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

Step 1.4.1.3

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

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

Step 1.4.2

Find the intersection of $x < - \frac{3}{2} or x \geq 1$ and $(- \infty, - \frac{3}{2}) \cup (- \frac{3}{2}, \infty)$ .

$x < - \frac{3}{2} or x \geq 1$

Step 1.5

To find the interval for the second piece, find where the inside of the absolute value is negative.

$\frac{x - 1}{2 x + 3} < 0$

Step 1.6

Solve the inequality.

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Step 1.6.1

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

$x - 1 = 0$

$2 x + 3 = 0$

Step 1.6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 1.6.3

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 1.6.4

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

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Step 1.6.4.1

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

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

Step 1.6.4.2

Simplify the left side.

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Step 1.6.4.2.1

Cancel the common factor of $2$ .

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Step 1.6.4.2.1.1

Cancel the common factor.

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

Step 1.6.4.2.1.2

Divide $x$ by $1$ .

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

Step 1.6.4.3

Simplify the right side.

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Step 1.6.4.3.1

Move the negative in front of the fraction.

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

Step 1.6.5

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

$x = 1$

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

Step 1.6.6

Consolidate the solutions.

$x = 1, - \frac{3}{2}$

Step 1.6.7

Find the domain of $| \frac{x - 1}{2 x + 3} | - 1$ .

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Step 1.6.7.1

Set the denominator in $\frac{x - 1}{2 x + 3}$ equal to $0$ to find where the expression is undefined.

$2 x + 3 = 0$

Step 1.6.7.2

Solve for $x$ .

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Step 1.6.7.2.1

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 1.6.7.2.2

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

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Step 1.6.7.2.2.1

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

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

Step 1.6.7.2.2.2

Simplify the left side.

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Step 1.6.7.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.6.7.2.2.2.1.1

Cancel the common factor.

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

Step 1.6.7.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.6.7.2.2.3

Simplify the right side.

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Step 1.6.7.2.2.3.1

Move the negative in front of the fraction.

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

Step 1.6.7.3

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

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

Step 1.6.8

Use each root to create test intervals.

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

$- \frac{3}{2} < x < 1$

$x > 1$

Step 1.6.9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 1.6.9.1

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

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Step 1.6.9.1.1

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

$x = - 4$

Step 1.6.9.1.2

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

$\frac{(- 4) - 1}{2 (- 4) + 3} < 0$

Step 1.6.9.1.3

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

False

Step 1.6.9.2

Test a value on the interval $- \frac{3}{2} < x < 1$ to see if it makes the inequality true.

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Step 1.6.9.2.1

Choose a value on the interval $- \frac{3}{2} < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 1.6.9.2.2

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

$\frac{(0) - 1}{2 (0) + 3} < 0$

Step 1.6.9.2.3

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

True

Step 1.6.9.3

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

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Step 1.6.9.3.1

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

$x = 4$

Step 1.6.9.3.2

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

$\frac{(4) - 1}{2 (4) + 3} < 0$

Step 1.6.9.3.3

The left side $0.\overline{27}$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 1.6.9.4

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

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

$- \frac{3}{2} < x < 1$ True

$x > 1$ False

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

$- \frac{3}{2} < x < 1$ True

$x > 1$ False

Step 1.6.10

The solution consists of all of the true intervals.

$- \frac{3}{2} < x < 1$

Step 1.7

In the piece where $\frac{x - 1}{2 x + 3}$ is negative, remove the absolute value and multiply by $- 1$ .

$- \frac{x - 1}{2 x + 3} < 1$

Step 1.8

Find the domain of $- \frac{x - 1}{2 x + 3} < 1$ and find the intersection with $- \frac{3}{2} < x < 1$ .

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Step 1.8.1

Find the domain of $- \frac{x - 1}{2 x + 3} < 1$ .

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Step 1.8.1.1

Set the denominator in $\frac{x - 1}{2 x + 3}$ equal to $0$ to find where the expression is undefined.

$2 x + 3 = 0$

Step 1.8.1.2

Solve for $x$ .

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Step 1.8.1.2.1

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 1.8.1.2.2

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

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Step 1.8.1.2.2.1

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

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

Step 1.8.1.2.2.2

Simplify the left side.

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Step 1.8.1.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.8.1.2.2.2.1.1

Cancel the common factor.

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

Step 1.8.1.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.8.1.2.2.3

Simplify the right side.

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Step 1.8.1.2.2.3.1

Move the negative in front of the fraction.

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

Step 1.8.1.3

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

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

Step 1.8.2

Find the intersection of $- \frac{3}{2} < x < 1$ and $(- \infty, - \frac{3}{2}) \cup (- \frac{3}{2}, \infty)$ .

$- \frac{3}{2} < x < 1$

Step 1.9

Write as a piecewise.

${\begin{cases} \frac{x - 1}{2 x + 3} < 1 & x < - \frac{3}{2} or x \geq 1 \\ - \frac{x - 1}{2 x + 3} < 1 & - \frac{3}{2} < x < 1 \end{cases}$

Step 2

Solve $\frac{x - 1}{2 x + 3} < 1$ when $x < - \frac{3}{2} or x \geq 1$ .

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Step 2.1

Solve $\frac{x - 1}{2 x + 3} < 1$ for $x$ .

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Step 2.1.1

Subtract $1$ from both sides of the inequality.

$\frac{x - 1}{2 x + 3} - 1 < 0$

Step 2.1.2

Simplify $\frac{x - 1}{2 x + 3} - 1$ .

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Step 2.1.2.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2 x + 3}{2 x + 3}$ .

$\frac{x - 1}{2 x + 3} - 1 \cdot \frac{2 x + 3}{2 x + 3} < 0$

Step 2.1.2.2

Combine $- 1$ and $\frac{2 x + 3}{2 x + 3}$ .

$\frac{x - 1}{2 x + 3} + \frac{- (2 x + 3)}{2 x + 3} < 0$

Step 2.1.2.3

Combine the numerators over the common denominator.

$\frac{x - 1 - (2 x + 3)}{2 x + 3} < 0$

Step 2.1.2.4

Simplify the numerator.

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Step 2.1.2.4.1

Apply the distributive property.

$\frac{x - 1 - (2 x) - 1 \cdot 3}{2 x + 3} < 0$

Step 2.1.2.4.2

Multiply $2$ by $- 1$ .

$\frac{x - 1 - 2 x - 1 \cdot 3}{2 x + 3} < 0$

Step 2.1.2.4.3

Multiply $- 1$ by $3$ .

$\frac{x - 1 - 2 x - 3}{2 x + 3} < 0$

Step 2.1.2.4.4

Subtract $2 x$ from $x$ .

$\frac{- x - 1 - 3}{2 x + 3} < 0$

Step 2.1.2.4.5

Subtract $3$ from $- 1$ .

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

Step 2.1.2.5

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

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

Step 2.1.2.6

Rewrite $- 4$ as $- 1 (4)$ .

$\frac{- (x) - 1 (4)}{2 x + 3} < 0$

Step 2.1.2.7

Factor $- 1$ out of $- (x) - 1 (4)$ .

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

Step 2.1.2.8

Rewrite $- (x + 4)$ as $- 1 (x + 4)$ .

$\frac{- 1 (x + 4)}{2 x + 3} < 0$

Step 2.1.2.9

Move the negative in front of the fraction.

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

Step 2.1.3

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

$x + 4 = 0$

$2 x + 3 = 0$

Step 2.1.4

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.1.5

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 2.1.6

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

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Step 2.1.6.1

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

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

Step 2.1.6.2

Simplify the left side.

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Step 2.1.6.2.1

Cancel the common factor of $2$ .

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Step 2.1.6.2.1.1

Cancel the common factor.

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

Step 2.1.6.2.1.2

Divide $x$ by $1$ .

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

Step 2.1.6.3

Simplify the right side.

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Step 2.1.6.3.1

Move the negative in front of the fraction.

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

Step 2.1.7

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

$x = - 4$

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

Step 2.1.8

Consolidate the solutions.

$x = - 4, - \frac{3}{2}$

Step 2.1.9

Find the domain of $- \frac{x + 4}{2 x + 3}$ .

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Step 2.1.9.1

Set the denominator in $\frac{x + 4}{2 x + 3}$ equal to $0$ to find where the expression is undefined.

$2 x + 3 = 0$

Step 2.1.9.2

Solve for $x$ .

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Step 2.1.9.2.1

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 2.1.9.2.2

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

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Step 2.1.9.2.2.1

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

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

Step 2.1.9.2.2.2

Simplify the left side.

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Step 2.1.9.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.1.9.2.2.2.1.1

Cancel the common factor.

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

Step 2.1.9.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.1.9.2.2.3

Simplify the right side.

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Step 2.1.9.2.2.3.1

Move the negative in front of the fraction.

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

Step 2.1.9.3

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

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

Step 2.1.10

Use each root to create test intervals.

$x < - 4$

$- 4 < x < - \frac{3}{2}$

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

Step 2.1.11

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.1.11.1

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

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Step 2.1.11.1.1

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

$x = - 6$

Step 2.1.11.1.2

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

$\frac{(- 6) - 1}{2 (- 6) + 3} < 1$

Step 2.1.11.1.3

The left side $0.\overline{7}$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 2.1.11.2

Test a value on the interval $- 4 < x < - \frac{3}{2}$ to see if it makes the inequality true.

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Step 2.1.11.2.1

Choose a value on the interval $- 4 < x < - \frac{3}{2}$ and see if this value makes the original inequality true.

$x = - 3$

Step 2.1.11.2.2

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

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

Step 2.1.11.2.3

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

False

Step 2.1.11.3

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

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Step 2.1.11.3.1

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

$x = 0$

Step 2.1.11.3.2

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

$\frac{(0) - 1}{2 (0) + 3} < 1$

Step 2.1.11.3.3

The left side $- 0.\overline{3}$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 2.1.11.4

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

$x < - 4$ True

$- 4 < x < - \frac{3}{2}$ False

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

$x < - 4$ True

$- 4 < x < - \frac{3}{2}$ False

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

Step 2.1.12

The solution consists of all of the true intervals.

$x < - 4$ or $x > - \frac{3}{2}$

Step 2.2

Find the intersection of $x < - 4 or x > - \frac{3}{2}$ and $x < - \frac{3}{2} or x \geq 1$ .

$x < - 4$ or $x \geq 1$

Step 3

Solve $- \frac{x - 1}{2 x + 3} < 1$ when $- \frac{3}{2} < x < 1$ .

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Step 3.1

Solve $- \frac{x - 1}{2 x + 3} < 1$ for $x$ .

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Step 3.1.1

Subtract $1$ from both sides of the inequality.

$- \frac{x - 1}{2 x + 3} - 1 < 0$

Step 3.1.2

Simplify $- \frac{x - 1}{2 x + 3} - 1$ .

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Step 3.1.2.1

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2 x + 3}{2 x + 3}$ .

$- \frac{x - 1}{2 x + 3} - 1 \cdot \frac{2 x + 3}{2 x + 3} < 0$

Step 3.1.2.2

Combine $- 1$ and $\frac{2 x + 3}{2 x + 3}$ .

$- \frac{x - 1}{2 x + 3} + \frac{- (2 x + 3)}{2 x + 3} < 0$

Step 3.1.2.3

Combine the numerators over the common denominator.

$\frac{- (x - 1) - (2 x + 3)}{2 x + 3} < 0$

Step 3.1.2.4

Simplify the numerator.

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Step 3.1.2.4.1

Factor $- 1$ out of $- (x - 1) - (2 x + 3)$ .

$\frac{- (x - 1 + 2 x + 3)}{2 x + 3} < 0$

Step 3.1.2.4.2

Add $x$ and $2 x$ .

$\frac{- (3 x - 1 + 3)}{2 x + 3} < 0$

Step 3.1.2.4.3

Add $- 1$ and $3$ .

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

Step 3.1.2.5

Move the negative in front of the fraction.

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

Step 3.1.3

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

$3 x + 2 = 0$

$2 x + 3 = 0$

Step 3.1.4

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Step 3.1.5

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

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Step 3.1.5.1

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

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

Step 3.1.5.2

Simplify the left side.

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Step 3.1.5.2.1

Cancel the common factor of $3$ .

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Step 3.1.5.2.1.1

Cancel the common factor.

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

Step 3.1.5.2.1.2

Divide $x$ by $1$ .

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

Step 3.1.5.3

Simplify the right side.

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Step 3.1.5.3.1

Move the negative in front of the fraction.

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

Step 3.1.6

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 3.1.7

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

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Step 3.1.7.1

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

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

Step 3.1.7.2

Simplify the left side.

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Step 3.1.7.2.1

Cancel the common factor of $2$ .

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Step 3.1.7.2.1.1

Cancel the common factor.

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

Step 3.1.7.2.1.2

Divide $x$ by $1$ .

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

Step 3.1.7.3

Simplify the right side.

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Step 3.1.7.3.1

Move the negative in front of the fraction.

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

Step 3.1.8

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

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

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

Step 3.1.9

Consolidate the solutions.

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

Step 3.1.10

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

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Step 3.1.10.1

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

$2 x + 3 = 0$

Step 3.1.10.2

Solve for $x$ .

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Step 3.1.10.2.1

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 3.1.10.2.2

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

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Step 3.1.10.2.2.1

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

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

Step 3.1.10.2.2.2

Simplify the left side.

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Step 3.1.10.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.1.10.2.2.2.1.1

Cancel the common factor.

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

Step 3.1.10.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 3.1.10.2.2.3

Simplify the right side.

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Step 3.1.10.2.2.3.1

Move the negative in front of the fraction.

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

Step 3.1.10.3

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

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

Step 3.1.11

Use each root to create test intervals.

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

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

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

Step 3.1.12

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 3.1.12.1

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

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Step 3.1.12.1.1

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

$x = - 4$

Step 3.1.12.1.2

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

$- \frac{(- 4) - 1}{2 (- 4) + 3} < 1$

Step 3.1.12.1.3

The left side $- 1$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 3.1.12.2

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

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Step 3.1.12.2.1

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

$x = - 1$

Step 3.1.12.2.2

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

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

Step 3.1.12.2.3

The left side $2$ is not less than the right side $1$ , which means that the given statement is false.

False

Step 3.1.12.3

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

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Step 3.1.12.3.1

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

$x = 0$

Step 3.1.12.3.2

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

$- \frac{(0) - 1}{2 (0) + 3} < 1$

Step 3.1.12.3.3

The left side $0.\overline{3}$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 3.1.12.4

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

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

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

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

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

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

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

Step 3.1.13

The solution consists of all of the true intervals.

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

Step 3.2

Find the intersection of $x < - \frac{3}{2} or x > - \frac{2}{3}$ and $- \frac{3}{2} < x < 1$ .

$- \frac{2}{3} < x < 1$

Step 4

Find the union of the solutions.

$x < - 4$ or $x > - \frac{2}{3}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x < - 4 or x > - \frac{2}{3}$

Interval Notation:

$(- \infty, - 4) \cup (- \frac{2}{3}, \infty)$

Step 6