Pre-Algebra Examples

$| \frac{- 1}{2 + x} | \leq \frac{1}{6}$

Step 1

Write $| \frac{- 1}{2 + x} | \leq \frac{1}{6}$ 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.

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

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.

$- 1 = 0$

$2 + x = 0$

Step 1.2.2

Since $- 1 \neq 0$ , there are no solutions.

No solution

Step 1.2.3

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 1.2.4

Find the domain of $\frac{6 | \frac{1}{2 + x} | - 1}{6}$ .

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Step 1.2.4.1

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

$2 + x = 0$

Step 1.2.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 1.2.4.3

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

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

Step 1.2.5

The solution consists of all of the true intervals.

$x < - 2$

Step 1.3

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

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

Step 1.4

Find the domain of $\frac{- 1}{2 + x} \leq \frac{1}{6}$ and find the intersection with $x < - 2$ .

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Step 1.4.1

Find the domain of $\frac{- 1}{2 + x} \leq \frac{1}{6}$ .

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Step 1.4.1.1

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

$2 + x = 0$

Step 1.4.1.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 1.4.1.3

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

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

Step 1.4.2

Find the intersection of $x < - 2$ and $(- \infty, - 2) \cup (- 2, \infty)$ .

$x < - 2$

Step 1.5

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

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

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.

$- 1 = 0$

$2 + x = 0$

Step 1.6.2

Since $- 1 \neq 0$ , there are no solutions.

No solution

Step 1.6.3

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 1.6.4

Find the domain of $\frac{6 | \frac{1}{2 + x} | - 1}{6}$ .

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Step 1.6.4.1

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

$2 + x = 0$

Step 1.6.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 1.6.4.3

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

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

Step 1.6.5

The solution consists of all of the true intervals.

$x > - 2$

Step 1.7

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

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

Step 1.8

Find the domain of $- \frac{- 1}{2 + x} \leq \frac{1}{6}$ and find the intersection with $x > - 2$ .

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Step 1.8.1

Find the domain of $- \frac{- 1}{2 + x} \leq \frac{1}{6}$ .

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Step 1.8.1.1

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

$2 + x = 0$

Step 1.8.1.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 1.8.1.3

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

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

Step 1.8.2

Find the intersection of $x > - 2$ and $(- \infty, - 2) \cup (- 2, \infty)$ .

$x > - 2$

Step 1.9

Write as a piecewise.

${\begin{cases} \frac{- 1}{2 + x} \leq \frac{1}{6} & x < - 2 \\ - \frac{- 1}{2 + x} \leq \frac{1}{6} & x > - 2 \end{cases}$

Step 1.10

Move the negative in front of the fraction.

${\begin{cases} - \frac{1}{2 + x} \leq \frac{1}{6} & x < - 2 \\ - \frac{- 1}{2 + x} \leq \frac{1}{6} & x > - 2 \end{cases}$

Step 1.11

Simplify $- \frac{- 1}{2 + x} \leq \frac{1}{6}$ .

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Step 1.11.1

Move the negative in front of the fraction.

${\begin{cases} - \frac{1}{2 + x} \leq \frac{1}{6} & x < - 2 \\ - - \frac{1}{2 + x} \leq \frac{1}{6} & x > - 2 \end{cases}$

Step 1.11.2

Multiply $- - \frac{1}{2 + x}$ .

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Step 1.11.2.1

Multiply $- 1$ by $- 1$ .

${\begin{cases} - \frac{1}{2 + x} \leq \frac{1}{6} & x < - 2 \\ 1 \frac{1}{2 + x} \leq \frac{1}{6} & x > - 2 \end{cases}$

Step 1.11.2.2

Multiply $\frac{1}{2 + x}$ by $1$ .

${\begin{cases} - \frac{1}{2 + x} \leq \frac{1}{6} & x < - 2 \\ \frac{1}{2 + x} \leq \frac{1}{6} & x > - 2 \end{cases}$

Step 2

Solve $- \frac{1}{2 + x} \leq \frac{1}{6}$ when $x < - 2$ .

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Step 2.1

Solve $- \frac{1}{2 + x} \leq \frac{1}{6}$ for $x$ .

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Step 2.1.1

Subtract $\frac{1}{6}$ from both sides of the inequality.

$- \frac{1}{2 + x} - \frac{1}{6} \leq 0$

Step 2.1.2

Simplify $- \frac{1}{2 + x} - \frac{1}{6}$ .

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Step 2.1.2.1

To write $- \frac{1}{2 + x}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$- \frac{1}{2 + x} \cdot \frac{6}{6} - \frac{1}{6} \leq 0$

Step 2.1.2.2

To write $- \frac{1}{6}$ as a fraction with a common denominator, multiply by $\frac{2 + x}{2 + x}$ .

$- \frac{1}{2 + x} \cdot \frac{6}{6} - \frac{1}{6} \cdot \frac{2 + x}{2 + x} \leq 0$

Step 2.1.2.3

Write each expression with a common denominator of $(2 + x) \cdot 6$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.1.2.3.1

Multiply $\frac{1}{2 + x}$ by $\frac{6}{6}$ .

$- \frac{6}{(2 + x) \cdot 6} - \frac{1}{6} \cdot \frac{2 + x}{2 + x} \leq 0$

Step 2.1.2.3.2

Multiply $\frac{1}{6}$ by $\frac{2 + x}{2 + x}$ .

$- \frac{6}{(2 + x) \cdot 6} - \frac{2 + x}{6 (2 + x)} \leq 0$

Step 2.1.2.3.3

Reorder the factors of $(2 + x) \cdot 6$ .

$- \frac{6}{6 (2 + x)} - \frac{2 + x}{6 (2 + x)} \leq 0$

Step 2.1.2.4

Combine the numerators over the common denominator.

$\frac{- 6 - (2 + x)}{6 (2 + x)} \leq 0$

Step 2.1.2.5

Simplify the numerator.

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Step 2.1.2.5.1

Factor $- 1$ out of $- 6 - (2 + x)$ .

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Step 2.1.2.5.1.1

Reorder the expression.

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Step 2.1.2.5.1.1.1

Reorder $2$ and $x$ .

$\frac{- 6 - (x + 2)}{6 (2 + x)} \leq 0$

Step 2.1.2.5.1.1.2

Reorder $- 6$ and $- (x + 2)$ .

$\frac{- (x + 2) - 6}{6 (2 + x)} \leq 0$

Step 2.1.2.5.1.2

Rewrite $- 6$ as $- 1 (6)$ .

$\frac{- (x + 2) - 1 (6)}{6 (2 + x)} \leq 0$

Step 2.1.2.5.1.3

Factor $- 1$ out of $- (x + 2) - 1 (6)$ .

$\frac{- (x + 2 + 6)}{6 (2 + x)} \leq 0$

Step 2.1.2.5.2

Add $2$ and $6$ .

$\frac{- (x + 8)}{6 (2 + x)} \leq 0$

Step 2.1.2.6

Move the negative in front of the fraction.

$- \frac{x + 8}{6 (2 + x)} \leq 0$

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.

$x + 8 = 0$

$2 + x = 0$

Step 2.1.4

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 2.1.5

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.1.6

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

$x = - 8$

$x = - 2$

Step 2.1.7

Consolidate the solutions.

$x = - 8, - 2$

Step 2.1.8

Find the domain of $- \frac{x + 8}{6 (2 + x)}$ .

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Step 2.1.8.1

Set the denominator in $\frac{x + 8}{6 (2 + x)}$ equal to $0$ to find where the expression is undefined.

$6 (2 + x) = 0$

Step 2.1.8.2

Solve for $x$ .

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Step 2.1.8.2.1

Divide each term in $6 (2 + x) = 0$ by $6$ and simplify.

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Step 2.1.8.2.1.1

Divide each term in $6 (2 + x) = 0$ by $6$ .

$\frac{6 (2 + x)}{6} = \frac{0}{6}$

Step 2.1.8.2.1.2

Simplify the left side.

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Step 2.1.8.2.1.2.1

Cancel the common factor of $6$ .

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Step 2.1.8.2.1.2.1.1

Cancel the common factor.

$\frac{6 (2 + x)}{6} = \frac{0}{6}$

Step 2.1.8.2.1.2.1.2

Divide $2 + x$ by $1$ .

$2 + x = \frac{0}{6}$

Step 2.1.8.2.1.3

Simplify the right side.

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Step 2.1.8.2.1.3.1

Divide $0$ by $6$ .

$2 + x = 0$

Step 2.1.8.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.1.8.3

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

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

Step 2.1.9

Use each root to create test intervals.

$x < - 8$

$- 8 < x < - 2$

$x > - 2$

Step 2.1.10

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 2.1.10.1

Test a value on the interval $x < - 8$ to see if it makes the inequality true.

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Step 2.1.10.1.1

Choose a value on the interval $x < - 8$ and see if this value makes the original inequality true.

$x = - 10$

Step 2.1.10.1.2

Replace $x$ with $- 10$ in the original inequality.

$- \frac{1}{2 - 10} \leq \frac{1}{6}$

Step 2.1.10.1.3

The left side $0.125$ is less than the right side $0.1\overline{6}$ , which means that the given statement is always true.

True

Step 2.1.10.2

Test a value on the interval $- 8 < x < - 2$ to see if it makes the inequality true.

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Step 2.1.10.2.1

Choose a value on the interval $- 8 < x < - 2$ and see if this value makes the original inequality true.

$x = - 5$

Step 2.1.10.2.2

Replace $x$ with $- 5$ in the original inequality.

$- \frac{1}{2 - 5} \leq \frac{1}{6}$

Step 2.1.10.2.3

The left side $0.\overline{3}$ is greater than the right side $0.1\overline{6}$ , which means that the given statement is false.

False

Step 2.1.10.3

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

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Step 2.1.10.3.1

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

$x = 0$

Step 2.1.10.3.2

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

$- \frac{1}{2 + 0} \leq \frac{1}{6}$

Step 2.1.10.3.3

The left side $- 0.5$ is less than the right side $0.1\overline{6}$ , which means that the given statement is always true.

True

Step 2.1.10.4

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

$x < - 8$ True

$- 8 < x < - 2$ False

$x > - 2$ True

$x < - 8$ True

$- 8 < x < - 2$ False

$x > - 2$ True

Step 2.1.11

The solution consists of all of the true intervals.

$x \leq - 8$ or $x > - 2$

Step 2.2

Find the intersection of $x \leq - 8 or x > - 2$ and $x < - 2$ .

$x \leq - 8$

Step 3

Solve $\frac{1}{2 + x} \leq \frac{1}{6}$ when $x > - 2$ .

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Step 3.1

Solve $\frac{1}{2 + x} \leq \frac{1}{6}$ for $x$ .

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Step 3.1.1

Subtract $\frac{1}{6}$ from both sides of the inequality.

$\frac{1}{2 + x} - \frac{1}{6} \leq 0$

Step 3.1.2

Simplify $\frac{1}{2 + x} - \frac{1}{6}$ .

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Step 3.1.2.1

To write $\frac{1}{2 + x}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$\frac{1}{2 + x} \cdot \frac{6}{6} - \frac{1}{6} \leq 0$

Step 3.1.2.2

To write $- \frac{1}{6}$ as a fraction with a common denominator, multiply by $\frac{2 + x}{2 + x}$ .

$\frac{1}{2 + x} \cdot \frac{6}{6} - \frac{1}{6} \cdot \frac{2 + x}{2 + x} \leq 0$

Step 3.1.2.3

Write each expression with a common denominator of $(2 + x) \cdot 6$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.1.2.3.1

Multiply $\frac{1}{2 + x}$ by $\frac{6}{6}$ .

$\frac{6}{(2 + x) \cdot 6} - \frac{1}{6} \cdot \frac{2 + x}{2 + x} \leq 0$

Step 3.1.2.3.2

Multiply $\frac{1}{6}$ by $\frac{2 + x}{2 + x}$ .

$\frac{6}{(2 + x) \cdot 6} - \frac{2 + x}{6 (2 + x)} \leq 0$

Step 3.1.2.3.3

Reorder the factors of $(2 + x) \cdot 6$ .

$\frac{6}{6 (2 + x)} - \frac{2 + x}{6 (2 + x)} \leq 0$

Step 3.1.2.4

Combine the numerators over the common denominator.

$\frac{6 - (2 + x)}{6 (2 + x)} \leq 0$

Step 3.1.2.5

Simplify the numerator.

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Step 3.1.2.5.1

Apply the distributive property.

$\frac{6 - 1 \cdot 2 - x}{6 (2 + x)} \leq 0$

Step 3.1.2.5.2

Multiply $- 1$ by $2$ .

$\frac{6 - 2 - x}{6 (2 + x)} \leq 0$

Step 3.1.2.5.3

Subtract $2$ from $6$ .

$\frac{4 - x}{6 (2 + x)} \leq 0$

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.

$4 - x = 0$

$2 + x = 0$

Step 3.1.4

Subtract $4$ from both sides of the equation.

$- x = - 4$

Step 3.1.5

Divide each term in $- x = - 4$ by $- 1$ and simplify.

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Step 3.1.5.1

Divide each term in $- x = - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1}$

Step 3.1.5.2

Simplify the left side.

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Step 3.1.5.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1}$

Step 3.1.5.2.2

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1}$

Step 3.1.5.3

Simplify the right side.

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Step 3.1.5.3.1

Divide $- 4$ by $- 1$ .

$x = 4$

Step 3.1.6

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3.1.7

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

$x = 4$

$x = - 2$

Step 3.1.8

Consolidate the solutions.

$x = 4, - 2$

Step 3.1.9

Find the domain of $\frac{4 - x}{6 (2 + x)}$ .

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Step 3.1.9.1

Set the denominator in $\frac{4 - x}{6 (2 + x)}$ equal to $0$ to find where the expression is undefined.

$6 (2 + x) = 0$

Step 3.1.9.2

Solve for $x$ .

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Step 3.1.9.2.1

Divide each term in $6 (2 + x) = 0$ by $6$ and simplify.

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Step 3.1.9.2.1.1

Divide each term in $6 (2 + x) = 0$ by $6$ .

$\frac{6 (2 + x)}{6} = \frac{0}{6}$

Step 3.1.9.2.1.2

Simplify the left side.

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Step 3.1.9.2.1.2.1

Cancel the common factor of $6$ .

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Step 3.1.9.2.1.2.1.1

Cancel the common factor.

$\frac{6 (2 + x)}{6} = \frac{0}{6}$

Step 3.1.9.2.1.2.1.2

Divide $2 + x$ by $1$ .

$2 + x = \frac{0}{6}$

Step 3.1.9.2.1.3

Simplify the right side.

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Step 3.1.9.2.1.3.1

Divide $0$ by $6$ .

$2 + x = 0$

Step 3.1.9.2.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 3.1.9.3

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

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

Step 3.1.10

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 4$

$x > 4$

Step 3.1.11

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 3.1.11.1

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

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Step 3.1.11.1.1

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

$x = - 4$

Step 3.1.11.1.2

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

$\frac{1}{2 - 4} \leq \frac{1}{6}$

Step 3.1.11.1.3

The left side $- 0.5$ is less than the right side $0.1\overline{6}$ , which means that the given statement is always true.

True

Step 3.1.11.2

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

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Step 3.1.11.2.1

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

$x = 0$

Step 3.1.11.2.2

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

$\frac{1}{2 + 0} \leq \frac{1}{6}$

Step 3.1.11.2.3

The left side $0.5$ is greater than the right side $0.1\overline{6}$ , which means that the given statement is false.

False

Step 3.1.11.3

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

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Step 3.1.11.3.1

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

$x = 6$

Step 3.1.11.3.2

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

$\frac{1}{2 + 6} \leq \frac{1}{6}$

Step 3.1.11.3.3

The left side $0.125$ is less than the right side $0.1\overline{6}$ , which means that the given statement is always true.

True

Step 3.1.11.4

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

$x < - 2$ True

$- 2 < x < 4$ False

$x > 4$ True

$x < - 2$ True

$- 2 < x < 4$ False

$x > 4$ True

Step 3.1.12

The solution consists of all of the true intervals.

$x < - 2$ or $x \geq 4$

Step 3.2

Find the intersection of $x < - 2 or x \geq 4$ and $x > - 2$ .

$x \geq 4$

Step 4

Find the union of the solutions.

$x \leq - 8$ or $x \geq 4$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 8 or x \geq 4$

Interval Notation:

$(- \infty, - 8] \cup [4, \infty)$

Step 6