Solve the Inequality for x 1/(|2x-3|)>2

$\frac{1}{| 2 x - 3 |} > 2$

Subtract $2$ from both sides of the inequality.

$\frac{1}{| 2 x - 3 |} - 2 > 0$

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

$\frac{1}{| 2 x - 3 |} - 2 \cdot \frac{| 2 x - 3 |}{| 2 x - 3 |} > 0$

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

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

Combine the numerators over the common denominator.

$\frac{1 - 2 | 2 x - 3 |}{| 2 x - 3 |} > 0$

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

$1 - 2 | 2 x - 3 | = 0$

$| 2 x - 3 | = 0$

Subtract $1$ from both sides of the equation.

$- 2 | 2 x - 3 | = - 1$

Divide each term in $- 2 | 2 x - 3 | = - 1$ by $- 2$ and simplify.

Divide each term in $- 2 | 2 x - 3 | = - 1$ by $- 2$ .

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $| 2 x - 3 |$ by $1$ .

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

Simplify the right side.

Dividing two negative values results in a positive value.

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

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

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

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

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

Move all terms not containing $x$ to the right side of the equation.

Add $3$ to both sides of the equation.

$2 x = \frac{1}{2} + 3$

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

$2 x = \frac{1}{2} + 3 \cdot \frac{2}{2}$

Combine $3$ and $\frac{2}{2}$ .

$2 x = \frac{1}{2} + \frac{3 \cdot 2}{2}$

Combine the numerators over the common denominator.

$2 x = \frac{1 + 3 \cdot 2}{2}$

Simplify the numerator.

Multiply $3$ by $2$ .

$2 x = \frac{1 + 6}{2}$

Add $1$ and $6$ .

$2 x = \frac{7}{2}$

Divide each term in $2 x = \frac{7}{2}$ by $2$ and simplify.

Divide each term in $2 x = \frac{7}{2}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{7}{2}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{7}{2}}{2}$

Divide $x$ by $1$ .

$x = \frac{\frac{7}{2}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{7}{2} \cdot \frac{1}{2}$

Multiply $\frac{7}{2} \cdot \frac{1}{2}$ .

Multiply $\frac{7}{2}$ by $\frac{1}{2}$ .

$x = \frac{7}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{7}{4}$

Next, use the negative value of the $\pm$ to find the second solution.

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

Move all terms not containing $x$ to the right side of the equation.

Add $3$ to both sides of the equation.

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

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

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

Combine $3$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $3$ by $2$ .

$2 x = \frac{- 1 + 6}{2}$

Add $- 1$ and $6$ .

$2 x = \frac{5}{2}$

Divide each term in $2 x = \frac{5}{2}$ by $2$ and simplify.

Divide each term in $2 x = \frac{5}{2}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{5}{2}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{5}{2}}{2}$

Divide $x$ by $1$ .

$x = \frac{\frac{5}{2}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{5}{2} \cdot \frac{1}{2}$

Multiply $\frac{5}{2} \cdot \frac{1}{2}$ .

Multiply $\frac{5}{2}$ by $\frac{1}{2}$ .

$x = \frac{5}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{5}{4}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{7}{4}, \frac{5}{4}$

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$2 x - 3 = \pm 0$

Plus or minus $0$ is $0$ .

$2 x - 3 = 0$

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 $1$ .

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

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

$x = \frac{7}{4}, \frac{5}{4}$

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

Consolidate the solutions.

$x = \frac{7}{4}, \frac{5}{4}, \frac{3}{2}$

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

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

$| 2 x - 3 | = 0$

Solve for $x$ .

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$2 x - 3 = \pm 0$

Plus or minus $0$ is $0$ .

$2 x - 3 = 0$

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 $1$ .

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

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

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

Use each root to create test intervals.

$x < \frac{5}{4}$

$\frac{5}{4} < x < \frac{3}{2}$

$\frac{3}{2} < x < \frac{7}{4}$

$x > \frac{7}{4}$

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 $x < \frac{5}{4}$ to see if it makes the inequality true.

Choose a value on the interval $x < \frac{5}{4}$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{1}{| 2 (0) - 3 |} > 2$

The left side $0.\overline{3}$ is not greater than the right side $2$ , which means that the given statement is false.

False

Test a value on the interval $\frac{5}{4} < x < \frac{3}{2}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{5}{4} < x < \frac{3}{2}$ and see if this value makes the original inequality true.

$x = 1.38$

Replace $x$ with $1.38$ in the original inequality.

$\frac{1}{| 2 (1.38) - 3 |} > 2$

The left side $4.1\overline{6}$ is greater than the right side $2$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{3}{2} < x < \frac{7}{4}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{3}{2} < x < \frac{7}{4}$ and see if this value makes the original inequality true.

$x = 1.62$

Replace $x$ with $1.62$ in the original inequality.

$\frac{1}{| 2 (1.62) - 3 |} > 2$

The left side $4.1\overline{6}$ is greater than the right side $2$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{7}{4}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{7}{4}$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{1}{| 2 (4) - 3 |} > 2$

The left side $0.2$ is not greater than the right side $2$ , which means that the given statement is false.

False

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

$x < \frac{5}{4}$ False

$\frac{5}{4} < x < \frac{3}{2}$ True

$\frac{3}{2} < x < \frac{7}{4}$ True

$x > \frac{7}{4}$ False

$x < \frac{5}{4}$ False

$\frac{5}{4} < x < \frac{3}{2}$ True

$\frac{3}{2} < x < \frac{7}{4}$ True

$x > \frac{7}{4}$ False

The solution consists of all of the true intervals.

$\frac{5}{4} < x < \frac{3}{2}$ or $\frac{3}{2} < x < \frac{7}{4}$

The result can be shown in multiple forms.

Inequality Form:

$\frac{5}{4} < x < \frac{3}{2} or \frac{3}{2} < x < \frac{7}{4}$

Interval Notation:

$(\frac{5}{4}, \frac{3}{2}) \cup (\frac{3}{2}, \frac{7}{4})$

Solve the Inequality for x 1/(|2x-3|)>2

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