Algebra Examples

$| 12 - \frac{1}{6 x} |$

Step 1

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

$12 - \frac{1}{6 x} \geq 0$

Step 2

Solve the inequality.

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Step 2.1

Subtract $12$ from both sides of the inequality.

$- \frac{1}{6 x} \geq - 12$

Step 2.2

Multiply both sides by $6 x$ .

$- \frac{1}{6 x} (6 x) = - 12 (6 x)$

Step 2.3

Simplify.

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Step 2.3.1

Simplify the left side.

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Step 2.3.1.1

Cancel the common factor of $6 x$ .

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Step 2.3.1.1.1

Move the leading negative in $- \frac{1}{6 x}$ into the numerator.

$\frac{- 1}{6 x} (6 x) = - 12 (6 x)$

Step 2.3.1.1.2

Cancel the common factor.

$\frac{- 1}{6 x} (6 x) = - 12 (6 x)$

Step 2.3.1.1.3

Rewrite the expression.

$- 1 = - 12 (6 x)$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Multiply $6$ by $- 12$ .

$- 1 = - 72 x$

Step 2.4

Solve for $x$ .

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Step 2.4.1

Rewrite the equation as $- 72 x = - 1$ .

$- 72 x = - 1$

Step 2.4.2

Divide each term in $- 72 x = - 1$ by $- 72$ and simplify.

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Step 2.4.2.1

Divide each term in $- 72 x = - 1$ by $- 72$ .

$\frac{- 72 x}{- 72} = \frac{- 1}{- 72}$

Step 2.4.2.2

Simplify the left side.

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Step 2.4.2.2.1

Cancel the common factor of $- 72$ .

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Step 2.4.2.2.1.1

Cancel the common factor.

$\frac{- 72 x}{- 72} = \frac{- 1}{- 72}$

Step 2.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 72}$

Step 2.4.2.3

Simplify the right side.

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Step 2.4.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{1}{72}$

Step 2.5

Find the domain of $12 - \frac{1}{6 x}$ .

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Step 2.5.1

Set the denominator in $\frac{1}{6 x}$ equal to $0$ to find where the expression is undefined.

$6 x = 0$

Step 2.5.2

Divide each term in $6 x = 0$ by $6$ and simplify.

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Step 2.5.2.1

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Step 2.5.2.2

Simplify the left side.

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Step 2.5.2.2.1

Cancel the common factor of $6$ .

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Step 2.5.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Step 2.5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Step 2.5.2.3

Simplify the right side.

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Step 2.5.2.3.1

Divide $0$ by $6$ .

$x = 0$

Step 2.5.3

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

$(- \infty, 0) \cup (0, \infty)$

Step 2.6

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{1}{72}$

$x > \frac{1}{72}$

Step 2.7

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 2.7.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 2.7.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 2.7.1.2

Replace $x$ with $- 2$ in the original inequality.

$12 - \frac{1}{6 (- 2)} \geq 0$

Step 2.7.1.3

The left side $12.08\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.7.2

Test a value on the interval $0 < x < \frac{1}{72}$ to see if it makes the inequality true.

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Step 2.7.2.1

Choose a value on the interval $0 < x < \frac{1}{72}$ and see if this value makes the original inequality true.

$x = 0.01$

Step 2.7.2.2

Replace $x$ with $0.01$ in the original inequality.

$12 - \frac{1}{6 (0.01)} \geq 0$

Step 2.7.2.3

The left side $- 4.\overline{6}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.7.3

Test a value on the interval $x > \frac{1}{72}$ to see if it makes the inequality true.

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Step 2.7.3.1

Choose a value on the interval $x > \frac{1}{72}$ and see if this value makes the original inequality true.

$x = 3$

Step 2.7.3.2

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

$12 - \frac{1}{6 (3)} \geq 0$

Step 2.7.3.3

The left side $11.9\overline{4}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.7.4

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

$x < 0$ True

$0 < x < \frac{1}{72}$ False

$x > \frac{1}{72}$ True

$x < 0$ True

$0 < x < \frac{1}{72}$ False

$x > \frac{1}{72}$ True

Step 2.8

The solution consists of all of the true intervals.

$x < 0$ or $x \geq \frac{1}{72}$

Step 3

In the piece where $12 - \frac{1}{6 x}$ is non-negative, remove the absolute value.

$12 - \frac{1}{6 x}$

Step 4

Find the domain of $12 - \frac{1}{6 x}$ and find the intersection with $x < 0 or x \geq \frac{1}{72}$ .

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Step 4.1

Find the domain of $12 - \frac{1}{6 x}$ .

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Step 4.1.1

Set the denominator in $\frac{1}{6 x}$ equal to $0$ to find where the expression is undefined.

$6 x = 0$

Step 4.1.2

Divide each term in $6 x = 0$ by $6$ and simplify.

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Step 4.1.2.1

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Step 4.1.2.2

Simplify the left side.

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Step 4.1.2.2.1

Cancel the common factor of $6$ .

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Step 4.1.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Step 4.1.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Step 4.1.2.3

Simplify the right side.

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Step 4.1.2.3.1

Divide $0$ by $6$ .

$x = 0$

Step 4.1.3

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

$(- \infty, 0) \cup (0, \infty)$

Step 4.2

Find the intersection of $x < 0 or x \geq \frac{1}{72}$ and $(- \infty, 0) \cup (0, \infty)$ .

$x < 0 or x \geq \frac{1}{72}$

Step 5

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

$12 - \frac{1}{6 x} < 0$

Step 6

Solve the inequality.

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Step 6.1

Subtract $12$ from both sides of the inequality.

$- \frac{1}{6 x} < - 12$

Step 6.2

Multiply both sides by $6 x$ .

$- \frac{1}{6 x} (6 x) = - 12 (6 x)$

Step 6.3

Simplify.

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Step 6.3.1

Simplify the left side.

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Step 6.3.1.1

Cancel the common factor of $6 x$ .

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Step 6.3.1.1.1

Move the leading negative in $- \frac{1}{6 x}$ into the numerator.

$\frac{- 1}{6 x} (6 x) = - 12 (6 x)$

Step 6.3.1.1.2

Cancel the common factor.

$\frac{- 1}{6 x} (6 x) = - 12 (6 x)$

Step 6.3.1.1.3

Rewrite the expression.

$- 1 = - 12 (6 x)$

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Multiply $6$ by $- 12$ .

$- 1 = - 72 x$

Step 6.4

Solve for $x$ .

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Step 6.4.1

Rewrite the equation as $- 72 x = - 1$ .

$- 72 x = - 1$

Step 6.4.2

Divide each term in $- 72 x = - 1$ by $- 72$ and simplify.

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Step 6.4.2.1

Divide each term in $- 72 x = - 1$ by $- 72$ .

$\frac{- 72 x}{- 72} = \frac{- 1}{- 72}$

Step 6.4.2.2

Simplify the left side.

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Step 6.4.2.2.1

Cancel the common factor of $- 72$ .

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Step 6.4.2.2.1.1

Cancel the common factor.

$\frac{- 72 x}{- 72} = \frac{- 1}{- 72}$

Step 6.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 72}$

Step 6.4.2.3

Simplify the right side.

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Step 6.4.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{1}{72}$

Step 6.5

Find the domain of $12 - \frac{1}{6 x}$ .

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Step 6.5.1

Set the denominator in $\frac{1}{6 x}$ equal to $0$ to find where the expression is undefined.

$6 x = 0$

Step 6.5.2

Divide each term in $6 x = 0$ by $6$ and simplify.

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Step 6.5.2.1

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Step 6.5.2.2

Simplify the left side.

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Step 6.5.2.2.1

Cancel the common factor of $6$ .

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Step 6.5.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Step 6.5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Step 6.5.2.3

Simplify the right side.

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Step 6.5.2.3.1

Divide $0$ by $6$ .

$x = 0$

Step 6.5.3

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

$(- \infty, 0) \cup (0, \infty)$

Step 6.6

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{1}{72}$

$x > \frac{1}{72}$

Step 6.7

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 6.7.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 6.7.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 6.7.1.2

Replace $x$ with $- 2$ in the original inequality.

$12 - \frac{1}{6 (- 2)} < 0$

Step 6.7.1.3

The left side $12.08\overline{3}$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 6.7.2

Test a value on the interval $0 < x < \frac{1}{72}$ to see if it makes the inequality true.

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Step 6.7.2.1

Choose a value on the interval $0 < x < \frac{1}{72}$ and see if this value makes the original inequality true.

$x = 0.01$

Step 6.7.2.2

Replace $x$ with $0.01$ in the original inequality.

$12 - \frac{1}{6 (0.01)} < 0$

Step 6.7.2.3

The left side $- 4.\overline{6}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 6.7.3

Test a value on the interval $x > \frac{1}{72}$ to see if it makes the inequality true.

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Step 6.7.3.1

Choose a value on the interval $x > \frac{1}{72}$ and see if this value makes the original inequality true.

$x = 3$

Step 6.7.3.2

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

$12 - \frac{1}{6 (3)} < 0$

Step 6.7.3.3

The left side $11.9\overline{4}$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 6.7.4

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

$x < 0$ False

$0 < x < \frac{1}{72}$ True

$x > \frac{1}{72}$ False

$x < 0$ False

$0 < x < \frac{1}{72}$ True

$x > \frac{1}{72}$ False

Step 6.8

The solution consists of all of the true intervals.

$0 < x < \frac{1}{72}$

Step 7

In the piece where $12 - \frac{1}{6 x}$ is negative, remove the absolute value and multiply by $- 1$ .

$- (12 - \frac{1}{6 x})$

Step 8

Find the domain of $- (12 - \frac{1}{6 x})$ and find the intersection with $0 < x < \frac{1}{72}$ .

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Step 8.1

Find the domain of $- (12 - \frac{1}{6 x})$ .

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Step 8.1.1

Set the denominator in $\frac{1}{6 x}$ equal to $0$ to find where the expression is undefined.

$6 x = 0$

Step 8.1.2

Divide each term in $6 x = 0$ by $6$ and simplify.

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Step 8.1.2.1

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Step 8.1.2.2

Simplify the left side.

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Step 8.1.2.2.1

Cancel the common factor of $6$ .

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Step 8.1.2.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Step 8.1.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Step 8.1.2.3

Simplify the right side.

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Step 8.1.2.3.1

Divide $0$ by $6$ .

$x = 0$

Step 8.1.3

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

$(- \infty, 0) \cup (0, \infty)$

Step 8.2

Find the intersection of $0 < x < \frac{1}{72}$ and $(- \infty, 0) \cup (0, \infty)$ .

$0 < x < \frac{1}{72}$

Step 9

Write as a piecewise.

${\begin{cases} 12 - \frac{1}{6 x} & x < 0 or x \geq \frac{1}{72} \\ - (12 - \frac{1}{6 x}) & 0 < x < \frac{1}{72} \end{cases}$

Step 10

Simplify $- (12 - \frac{1}{6 x})$ .

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Step 10.1

Apply the distributive property.

${\begin{cases} 12 - \frac{1}{6 x} & x < 0 or x \geq \frac{1}{72} \\ - 1 \cdot 12 - - \frac{1}{6 x} & 0 < x < \frac{1}{72} \end{cases}$

Step 10.2

Multiply $- 1$ by $12$ .

${\begin{cases} 12 - \frac{1}{6 x} & x < 0 or x \geq \frac{1}{72} \\ - 12 - - \frac{1}{6 x} & 0 < x < \frac{1}{72} \end{cases}$

Step 10.3

Multiply $- - \frac{1}{6 x}$ .

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Step 10.3.1

Multiply $- 1$ by $- 1$ .

${\begin{cases} 12 - \frac{1}{6 x} & x < 0 or x \geq \frac{1}{72} \\ - 12 + 1 \frac{1}{6 x} & 0 < x < \frac{1}{72} \end{cases}$

Step 10.3.2

Multiply $\frac{1}{6 x}$ by $1$ .

${\begin{cases} 12 - \frac{1}{6 x} & x < 0 or x \geq \frac{1}{72} \\ - 12 + \frac{1}{6 x} & 0 < x < \frac{1}{72} \end{cases}$