Calculus Examples

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

Step 1

Subtract $1$ from both sides of the inequality.

$\frac{| 2 x - 1 |}{2 - x} - 1 < 0$

Step 2

Simplify $\frac{| 2 x - 1 |}{2 - x} - 1$ .

Tap for more steps...

Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Combine the numerators over the common denominator.

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

Step 2.4

Simplify the numerator.

Tap for more steps...

Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Multiply $- 1$ by $2$ .

$\frac{| 2 x - 1 | - 2 - - x}{2 - x} < 0$

Step 2.4.3

Multiply $- - x$ .

Tap for more steps...

Step 2.4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 2.4.3.2

Multiply $x$ by $1$ .

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

Step 2.4.4

Reorder terms.

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

Step 3

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

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

$2 - x = 0$

Step 4

Move all terms not containing $| 2 x - 1 |$ to the right side of the equation.

Tap for more steps...

Step 4.1

Subtract $x$ from both sides of the equation.

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

Step 4.2

Add $2$ to both sides of the equation.

$| 2 x - 1 | = - x + 2$

Step 5

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

$2 x - 1 = \pm (- x + 2)$

Step 6

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

Tap for more steps...

Step 6.1

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

$2 x - 1 = - x + 2$

Step 6.2

Move all terms containing $x$ to the left side of the equation.

Tap for more steps...

Step 6.2.1

Add $x$ to both sides of the equation.

$2 x - 1 + x = 2$

Step 6.2.2

Add $2 x$ and $x$ .

$3 x - 1 = 2$

Step 6.3

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

Tap for more steps...

Step 6.3.1

Add $1$ to both sides of the equation.

$3 x = 2 + 1$

Step 6.3.2

Add $2$ and $1$ .

$3 x = 3$

Step 6.4

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

Tap for more steps...

Step 6.4.1

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

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

Step 6.4.2

Simplify the left side.

Tap for more steps...

Step 6.4.2.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 6.4.2.1.1

Cancel the common factor.

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

Step 6.4.2.1.2

Divide $x$ by $1$ .

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

Step 6.4.3

Simplify the right side.

Tap for more steps...

Step 6.4.3.1

Divide $3$ by $3$ .

$x = 1$

Step 6.5

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

$2 x - 1 = - (- x + 2)$

Step 6.6

Simplify $- (- x + 2)$ .

Tap for more steps...

Step 6.6.1

Rewrite.

$2 x - 1 = 0 + 0 - (- x + 2)$

Step 6.6.2

Simplify by adding zeros.

$2 x - 1 = - (- x + 2)$

Step 6.6.3

Apply the distributive property.

$2 x - 1 = - - x - 1 \cdot 2$

Step 6.6.4

Multiply $- - x$ .

Tap for more steps...

Step 6.6.4.1

Multiply $- 1$ by $- 1$ .

$2 x - 1 = 1 x - 1 \cdot 2$

Step 6.6.4.2

Multiply $x$ by $1$ .

$2 x - 1 = x - 1 \cdot 2$

Step 6.6.5

Multiply $- 1$ by $2$ .

$2 x - 1 = x - 2$

Step 6.7

Move all terms containing $x$ to the left side of the equation.

Tap for more steps...

Step 6.7.1

Subtract $x$ from both sides of the equation.

$2 x - 1 - x = - 2$

Step 6.7.2

Subtract $x$ from $2 x$ .

$x - 1 = - 2$

Step 6.8

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

Tap for more steps...

Step 6.8.1

Add $1$ to both sides of the equation.

$x = - 2 + 1$

Step 6.8.2

Add $- 2$ and $1$ .

$x = - 1$

Step 6.9

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

$x = 1, - 1$

Step 7

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 8

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

Tap for more steps...

Step 8.1

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

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

Step 8.2

Simplify the left side.

Tap for more steps...

Step 8.2.1

Dividing two negative values results in a positive value.

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

Step 8.2.2

Divide $x$ by $1$ .

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

Step 8.3

Simplify the right side.

Tap for more steps...

Step 8.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 9

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

$x = 1, - 1$

$x = 2$

Step 10

Consolidate the solutions.

$x = 1, - 1, 2$

Step 11

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

Tap for more steps...

Step 11.1

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

$2 - x = 0$

Step 11.2

Solve for $x$ .

Tap for more steps...

Step 11.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 11.2.2

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

Tap for more steps...

Step 11.2.2.1

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

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

Step 11.2.2.2

Simplify the left side.

Tap for more steps...

Step 11.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 11.2.2.2.2

Divide $x$ by $1$ .

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

Step 11.2.2.3

Simplify the right side.

Tap for more steps...

Step 11.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 11.3

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

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

Step 12

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$1 < x < 2$

$x > 2$

Step 13

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

Tap for more steps...

Step 13.1

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

Tap for more steps...

Step 13.1.1

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

$x = - 4$

Step 13.1.2

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

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

Step 13.1.3

The left side $1.5$ is not less than the right side $1$ , which means that the given statement is false.

False

Step 13.2

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

Tap for more steps...

Step 13.2.1

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

$x = 0$

Step 13.2.2

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

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

Step 13.2.3

The left side $0.5$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 13.3

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

Tap for more steps...

Step 13.3.1

Choose a value on the interval $1 < x < 2$ and see if this value makes the original inequality true.

$x = 1.5$

Step 13.3.2

Replace $x$ with $1.5$ in the original inequality.

$\frac{| 2 (1.5) - 1 |}{2 - (1.5)} < 1$

Step 13.3.3

The left side $4$ is not less than the right side $1$ , which means that the given statement is false.

False

Step 13.4

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

Tap for more steps...

Step 13.4.1

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

$x = 4$

Step 13.4.2

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

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

Step 13.4.3

The left side $- 3.5$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 13.5

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

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 2$ False

$x > 2$ True

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 2$ False

$x > 2$ True

Step 14

The solution consists of all of the true intervals.

$- 1 < x < 1$ or $x > 2$

Step 15

The result can be shown in multiple forms.

Inequality Form:

$- 1 < x < 1 or x > 2$

Interval Notation:

$(- 1, 1) \cup (2, \infty)$

Step 16