Calculus Examples

$| x + 1 | - | x - 3 | > - 1$

Step 1

Replace $>$ with $=$ in $| x + 1 | - | x - 3 | > - 1$ .

$| x + 1 | - | x - 3 | = - 1$

Step 2

Solve for $| x - 3 |$ .

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Step 2.1

Subtract $| x + 1 |$ from both sides of the equation.

$- | x - 3 | = - 1 - | x + 1 |$

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

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

Step 2.2.2.2

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

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

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Simplify each term.

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Step 2.2.3.1.1

Divide $- 1$ by $- 1$ .

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

Step 2.2.3.1.2

Dividing two negative values results in a positive value.

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

Step 2.2.3.1.3

Divide $| x + 1 |$ by $1$ .

$| x - 3 | = 1 + | x + 1 |$

Step 3

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

$x - 3 = \pm (1 + | x + 1 |)$

Step 4

The result consists of both the positive and negative portions of the $\pm$ .

$x - 3 = 1 + | x + 1 |$

$x - 3 = - (1 + | x + 1 |)$

Step 5

Solve $x - 3 = 1 + | x + 1 |$ for $x$ .

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Step 5.1

Solve for $| x + 1 |$ .

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Step 5.1.1

Rewrite the equation as $1 + | x + 1 | = x - 3$ .

$1 + | x + 1 | = x - 3$

Step 5.1.2

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

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Step 5.1.2.1

Subtract $1$ from both sides of the equation.

$| x + 1 | = x - 3 - 1$

Step 5.1.2.2

Subtract $1$ from $- 3$ .

$| x + 1 | = x - 4$

Step 5.2

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

$x + 1 = \pm (x - 4)$

Step 5.3

The result consists of both the positive and negative portions of the $\pm$ .

$x + 1 = x - 4$

$x + 1 = - (x - 4)$

Step 5.4

Solve $x + 1 = x - 4$ for $x$ .

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Step 5.4.1

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

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Step 5.4.1.1

Subtract $x$ from both sides of the equation.

$x + 1 - x = - 4$

Step 5.4.1.2

Combine the opposite terms in $x + 1 - x$ .

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Step 5.4.1.2.1

Subtract $x$ from $x$ .

$0 + 1 = - 4$

Step 5.4.1.2.2

Add $0$ and $1$ .

$1 = - 4$

Step 5.4.2

Since $1 \neq - 4$ , there are no solutions.

No solution

Step 5.5

Solve $x + 1 = - (x - 4)$ for $x$ .

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Step 5.5.1

Simplify $- (x - 4)$ .

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Step 5.5.1.1

Rewrite.

$x + 1 = 0 + 0 - (x - 4)$

Step 5.5.1.2

Simplify by adding zeros.

$x + 1 = - (x - 4)$

Step 5.5.1.3

Apply the distributive property.

$x + 1 = - x - - 4$

Step 5.5.1.4

Multiply $- 1$ by $- 4$ .

$x + 1 = - x + 4$

Step 5.5.2

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

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Step 5.5.2.1

Add $x$ to both sides of the equation.

$x + 1 + x = 4$

Step 5.5.2.2

Add $x$ and $x$ .

$2 x + 1 = 4$

Step 5.5.3

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

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Step 5.5.3.1

Subtract $1$ from both sides of the equation.

$2 x = 4 - 1$

Step 5.5.3.2

Subtract $1$ from $4$ .

$2 x = 3$

Step 5.5.4

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

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Step 5.5.4.1

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

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

Step 5.5.4.2

Simplify the left side.

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Step 5.5.4.2.1

Cancel the common factor of $2$ .

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Step 5.5.4.2.1.1

Cancel the common factor.

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

Step 5.5.4.2.1.2

Divide $x$ by $1$ .

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

Step 5.6

Consolidate the solutions.

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

Step 6

Solve $x - 3 = - (1 + | x + 1 |)$ for $x$ .

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Step 6.1

Solve for $| x + 1 |$ .

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Step 6.1.1

Rewrite the equation as $- (1 + | x + 1 |) = x - 3$ .

$- (1 + | x + 1 |) = x - 3$

Step 6.1.2

Simplify $- (1 + | x + 1 |)$ .

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Step 6.1.2.1

Apply the distributive property.

$- 1 \cdot 1 - | x + 1 | = x - 3$

Step 6.1.2.2

Multiply $- 1$ by $1$ .

$- 1 - | x + 1 | = x - 3$

Step 6.1.3

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

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Step 6.1.3.1

Add $1$ to both sides of the equation.

$- | x + 1 | = x - 3 + 1$

Step 6.1.3.2

Add $- 3$ and $1$ .

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

Step 6.1.4

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

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Step 6.1.4.1

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

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

Step 6.1.4.2

Simplify the left side.

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Step 6.1.4.2.1

Dividing two negative values results in a positive value.

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

Step 6.1.4.2.2

Divide $| x + 1 |$ by $1$ .

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

Step 6.1.4.3

Simplify the right side.

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Step 6.1.4.3.1

Simplify each term.

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Step 6.1.4.3.1.1

Move the negative one from the denominator of $\frac{x}{- 1}$ .

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

Step 6.1.4.3.1.2

Rewrite $- 1 \cdot x$ as $- x$ .

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

Step 6.1.4.3.1.3

Divide $- 2$ by $- 1$ .

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

Step 6.2

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

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

Step 6.3

The result consists of both the positive and negative portions of the $\pm$ .

$x + 1 = - x + 2$

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

Step 6.4

Solve $x + 1 = - x + 2$ for $x$ .

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Step 6.4.1

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

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Step 6.4.1.1

Add $x$ to both sides of the equation.

$x + 1 + x = 2$

Step 6.4.1.2

Add $x$ and $x$ .

$2 x + 1 = 2$

Step 6.4.2

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

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Step 6.4.2.1

Subtract $1$ from both sides of the equation.

$2 x = 2 - 1$

Step 6.4.2.2

Subtract $1$ from $2$ .

$2 x = 1$

Step 6.4.3

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

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Step 6.4.3.1

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

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

Step 6.4.3.2

Simplify the left side.

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Step 6.4.3.2.1

Cancel the common factor of $2$ .

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Step 6.4.3.2.1.1

Cancel the common factor.

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

Step 6.4.3.2.1.2

Divide $x$ by $1$ .

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

Step 6.5

Solve $x + 1 = - (- x + 2)$ for $x$ .

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Step 6.5.1

Simplify $- (- x + 2)$ .

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Step 6.5.1.1

Rewrite.

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

Step 6.5.1.2

Simplify by adding zeros.

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

Step 6.5.1.3

Apply the distributive property.

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

Step 6.5.1.4

Multiply $- - x$ .

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Step 6.5.1.4.1

Multiply $- 1$ by $- 1$ .

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

Step 6.5.1.4.2

Multiply $x$ by $1$ .

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

Step 6.5.1.5

Multiply $- 1$ by $2$ .

$x + 1 = x - 2$

Step 6.5.2

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

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Step 6.5.2.1

Subtract $x$ from both sides of the equation.

$x + 1 - x = - 2$

Step 6.5.2.2

Combine the opposite terms in $x + 1 - x$ .

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Step 6.5.2.2.1

Subtract $x$ from $x$ .

$0 + 1 = - 2$

Step 6.5.2.2.2

Add $0$ and $1$ .

$1 = - 2$

Step 6.5.3

Since $1 \neq - 2$ , there are no solutions.

No solution

Step 6.6

Consolidate the solutions.

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

Step 7

Consolidate the solutions.

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

Step 8

Use each root to create test intervals.

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

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

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

Step 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 9.1

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

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Step 9.1.1

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

$x = 0$

Step 9.1.2

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

$| (0) + 1 | - | (0) - 3 | > - 1$

Step 9.1.3

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

False

Step 9.2

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

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Step 9.2.1

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

$x = 1$

Step 9.2.2

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

$| (1) + 1 | - | (1) - 3 | > - 1$

Step 9.2.3

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

True

Step 9.3

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

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Step 9.3.1

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

$x = 4$

Step 9.3.2

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

$| (4) + 1 | - | (4) - 3 | > - 1$

Step 9.3.3

The left side $4$ is greater than the right side $- 1$ , which means that the given statement is always true.

True

Step 9.4

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

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

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

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

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

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

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

Step 10

The solution consists of all of the true intervals.

$\frac{1}{2} < x < \frac{3}{2}$ or $x > \frac{3}{2}$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$\frac{1}{2} < x < \frac{3}{2} or x > \frac{3}{2}$

Interval Notation:

$(\frac{1}{2}, \frac{3}{2}) \cup (\frac{3}{2}, \infty)$

Step 12