Pre-Algebra Examples

$| \frac{3 x - 1}{1 - 5 x} | > 2$

Step 1

Write $| \frac{3 x - 1}{1 - 5 x} | > 2$ 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.

$\frac{3 x - 1}{1 - 5 x} \geq 0$

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.

$3 x - 1 = 0$

$1 - 5 x = 0$

Step 1.2.2

Add $1$ to both sides of the equation.

$3 x = 1$

Step 1.2.3

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

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Step 1.2.3.1

Divide each term in $3 x = 1$ by $3$ .

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

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Cancel the common factor of $3$ .

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Step 1.2.3.2.1.1

Cancel the common factor.

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

Step 1.2.3.2.1.2

Divide $x$ by $1$ .

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

Step 1.2.4

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

Step 1.2.5

Divide each term in $- 5 x = - 1$ by $- 5$ and simplify.

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Step 1.2.5.1

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.2.5.2

Simplify the left side.

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Step 1.2.5.2.1

Cancel the common factor of $- 5$ .

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Step 1.2.5.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.2.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Step 1.2.5.3

Simplify the right side.

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Step 1.2.5.3.1

Dividing two negative values results in a positive value.

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

Step 1.2.6

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

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

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

Step 1.2.7

Consolidate the solutions.

$x = \frac{1}{3}, \frac{1}{5}$

Step 1.2.8

Find the domain of $| \frac{3 x - 1}{1 - 5 x} | - 2$ .

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Step 1.2.8.1

Set the denominator in $\frac{3 x - 1}{1 - 5 x}$ equal to $0$ to find where the expression is undefined.

$1 - 5 x = 0$

Step 1.2.8.2

Solve for $x$ .

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Step 1.2.8.2.1

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

Step 1.2.8.2.2

Divide each term in $- 5 x = - 1$ by $- 5$ and simplify.

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Step 1.2.8.2.2.1

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.2.8.2.2.2

Simplify the left side.

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Step 1.2.8.2.2.2.1

Cancel the common factor of $- 5$ .

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Step 1.2.8.2.2.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.2.8.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Step 1.2.8.2.2.3

Simplify the right side.

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Step 1.2.8.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 1.2.8.3

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

$(- \infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$

Step 1.2.9

Use each root to create test intervals.

$x < \frac{1}{5}$

$\frac{1}{5} < x < \frac{1}{3}$

$x > \frac{1}{3}$

Step 1.2.10

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 1.2.10.1

Test a value on the interval $x < \frac{1}{5}$ to see if it makes the inequality true.

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Step 1.2.10.1.1

Choose a value on the interval $x < \frac{1}{5}$ and see if this value makes the original inequality true.

$x = 0$

Step 1.2.10.1.2

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

$\frac{3 (0) - 1}{1 - 5 \cdot 0} \geq 0$

Step 1.2.10.1.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.10.2

Test a value on the interval $\frac{1}{5} < x < \frac{1}{3}$ to see if it makes the inequality true.

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Step 1.2.10.2.1

Choose a value on the interval $\frac{1}{5} < x < \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0.27$

Step 1.2.10.2.2

Replace $x$ with $0.27$ in the original inequality.

$\frac{3 (0.27) - 1}{1 - 5 \cdot 0.27} \geq 0$

Step 1.2.10.2.3

The left side $0.5\overline{428571}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.10.3

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

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Step 1.2.10.3.1

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

$x = 3$

Step 1.2.10.3.2

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

$\frac{3 (3) - 1}{1 - 5 \cdot 3} \geq 0$

Step 1.2.10.3.3

The left side $- 0.\overline{571428}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.10.4

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

$x < \frac{1}{5}$ False

$\frac{1}{5} < x < \frac{1}{3}$ True

$x > \frac{1}{3}$ False

$x < \frac{1}{5}$ False

$\frac{1}{5} < x < \frac{1}{3}$ True

$x > \frac{1}{3}$ False

Step 1.2.11

The solution consists of all of the true intervals.

$\frac{1}{5} < x \leq \frac{1}{3}$

Step 1.3

In the piece where $\frac{3 x - 1}{1 - 5 x}$ is non-negative, remove the absolute value.

$\frac{3 x - 1}{1 - 5 x} > 2$

Step 1.4

Find the domain of $\frac{3 x - 1}{1 - 5 x} > 2$ and find the intersection with $\frac{1}{5} < x \leq \frac{1}{3}$ .

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Step 1.4.1

Find the domain of $\frac{3 x - 1}{1 - 5 x} > 2$ .

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Step 1.4.1.1

Set the denominator in $\frac{3 x - 1}{1 - 5 x}$ equal to $0$ to find where the expression is undefined.

$1 - 5 x = 0$

Step 1.4.1.2

Solve for $x$ .

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Step 1.4.1.2.1

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

Step 1.4.1.2.2

Divide each term in $- 5 x = - 1$ by $- 5$ and simplify.

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Step 1.4.1.2.2.1

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.4.1.2.2.2

Simplify the left side.

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Step 1.4.1.2.2.2.1

Cancel the common factor of $- 5$ .

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Step 1.4.1.2.2.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.4.1.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Step 1.4.1.2.2.3

Simplify the right side.

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Step 1.4.1.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 1.4.1.3

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

$(- \infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$

Step 1.4.2

Find the intersection of $\frac{1}{5} < x \leq \frac{1}{3}$ and $(- \infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$ .

$\frac{1}{5} < x \leq \frac{1}{3}$

Step 1.5

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

$\frac{3 x - 1}{1 - 5 x} < 0$

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.

$3 x - 1 = 0$

$1 - 5 x = 0$

Step 1.6.2

Add $1$ to both sides of the equation.

$3 x = 1$

Step 1.6.3

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

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Step 1.6.3.1

Divide each term in $3 x = 1$ by $3$ .

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

Step 1.6.3.2

Simplify the left side.

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Step 1.6.3.2.1

Cancel the common factor of $3$ .

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Step 1.6.3.2.1.1

Cancel the common factor.

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

Step 1.6.3.2.1.2

Divide $x$ by $1$ .

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

Step 1.6.4

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

Step 1.6.5

Divide each term in $- 5 x = - 1$ by $- 5$ and simplify.

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Step 1.6.5.1

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.6.5.2

Simplify the left side.

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Step 1.6.5.2.1

Cancel the common factor of $- 5$ .

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Step 1.6.5.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.6.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Step 1.6.5.3

Simplify the right side.

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Step 1.6.5.3.1

Dividing two negative values results in a positive value.

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

Step 1.6.6

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

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

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

Step 1.6.7

Consolidate the solutions.

$x = \frac{1}{3}, \frac{1}{5}$

Step 1.6.8

Find the domain of $| \frac{3 x - 1}{1 - 5 x} | - 2$ .

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Step 1.6.8.1

Set the denominator in $\frac{3 x - 1}{1 - 5 x}$ equal to $0$ to find where the expression is undefined.

$1 - 5 x = 0$

Step 1.6.8.2

Solve for $x$ .

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Step 1.6.8.2.1

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

Step 1.6.8.2.2

Divide each term in $- 5 x = - 1$ by $- 5$ and simplify.

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Step 1.6.8.2.2.1

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.6.8.2.2.2

Simplify the left side.

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Step 1.6.8.2.2.2.1

Cancel the common factor of $- 5$ .

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Step 1.6.8.2.2.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.6.8.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Step 1.6.8.2.2.3

Simplify the right side.

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Step 1.6.8.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 1.6.8.3

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

$(- \infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$

Step 1.6.9

Use each root to create test intervals.

$x < \frac{1}{5}$

$\frac{1}{5} < x < \frac{1}{3}$

$x > \frac{1}{3}$

Step 1.6.10

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 1.6.10.1

Test a value on the interval $x < \frac{1}{5}$ to see if it makes the inequality true.

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Step 1.6.10.1.1

Choose a value on the interval $x < \frac{1}{5}$ and see if this value makes the original inequality true.

$x = 0$

Step 1.6.10.1.2

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

$\frac{3 (0) - 1}{1 - 5 \cdot 0} < 0$

Step 1.6.10.1.3

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

True

Step 1.6.10.2

Test a value on the interval $\frac{1}{5} < x < \frac{1}{3}$ to see if it makes the inequality true.

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Step 1.6.10.2.1

Choose a value on the interval $\frac{1}{5} < x < \frac{1}{3}$ and see if this value makes the original inequality true.

$x = 0.27$

Step 1.6.10.2.2

Replace $x$ with $0.27$ in the original inequality.

$\frac{3 (0.27) - 1}{1 - 5 \cdot 0.27} < 0$

Step 1.6.10.2.3

The left side $0.5\overline{428571}$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 1.6.10.3

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

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Step 1.6.10.3.1

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

$x = 3$

Step 1.6.10.3.2

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

$\frac{3 (3) - 1}{1 - 5 \cdot 3} < 0$

Step 1.6.10.3.3

The left side $- 0.\overline{571428}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 1.6.10.4

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

$x < \frac{1}{5}$ True

$\frac{1}{5} < x < \frac{1}{3}$ False

$x > \frac{1}{3}$ True

$x < \frac{1}{5}$ True

$\frac{1}{5} < x < \frac{1}{3}$ False

$x > \frac{1}{3}$ True

Step 1.6.11

The solution consists of all of the true intervals.

$x < \frac{1}{5}$ or $x > \frac{1}{3}$

Step 1.7

In the piece where $\frac{3 x - 1}{1 - 5 x}$ is negative, remove the absolute value and multiply by $- 1$ .

$- \frac{3 x - 1}{1 - 5 x} > 2$

Step 1.8

Find the domain of $- \frac{3 x - 1}{1 - 5 x} > 2$ and find the intersection with $x < \frac{1}{5} or x > \frac{1}{3}$ .

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Step 1.8.1

Find the domain of $- \frac{3 x - 1}{1 - 5 x} > 2$ .

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Step 1.8.1.1

Set the denominator in $\frac{3 x - 1}{1 - 5 x}$ equal to $0$ to find where the expression is undefined.

$1 - 5 x = 0$

Step 1.8.1.2

Solve for $x$ .

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Step 1.8.1.2.1

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

Step 1.8.1.2.2

Divide each term in $- 5 x = - 1$ by $- 5$ and simplify.

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Step 1.8.1.2.2.1

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.8.1.2.2.2

Simplify the left side.

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Step 1.8.1.2.2.2.1

Cancel the common factor of $- 5$ .

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Step 1.8.1.2.2.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 1.8.1.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Step 1.8.1.2.2.3

Simplify the right side.

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Step 1.8.1.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 1.8.1.3

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

$(- \infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$

Step 1.8.2

Find the intersection of $x < \frac{1}{5} or x > \frac{1}{3}$ and $(- \infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$ .

$x < \frac{1}{5} or x > \frac{1}{3}$

Step 1.9

Write as a piecewise.

${\begin{cases} \frac{3 x - 1}{1 - 5 x} > 2 & \frac{1}{5} < x \leq \frac{1}{3} \\ - \frac{3 x - 1}{1 - 5 x} > 2 & x < \frac{1}{5} or x > \frac{1}{3} \end{cases}$

Step 2

Solve $\frac{3 x - 1}{1 - 5 x} > 2$ for $x$ .

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Step 2.1

Subtract $2$ from both sides of the inequality.

$\frac{3 x - 1}{1 - 5 x} - 2 > 0$

Step 2.2

Simplify $\frac{3 x - 1}{1 - 5 x} - 2$ .

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Step 2.2.1

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{1 - 5 x}{1 - 5 x}$ .

$\frac{3 x - 1}{1 - 5 x} - 2 \cdot \frac{1 - 5 x}{1 - 5 x} > 0$

Step 2.2.2

Combine $- 2$ and $\frac{1 - 5 x}{1 - 5 x}$ .

$\frac{3 x - 1}{1 - 5 x} + \frac{- 2 (1 - 5 x)}{1 - 5 x} > 0$

Step 2.2.3

Combine the numerators over the common denominator.

$\frac{3 x - 1 - 2 (1 - 5 x)}{1 - 5 x} > 0$

Step 2.2.4

Simplify the numerator.

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Step 2.2.4.1

Apply the distributive property.

$\frac{3 x - 1 - 2 \cdot 1 - 2 (- 5 x)}{1 - 5 x} > 0$

Step 2.2.4.2

Multiply $- 2$ by $1$ .

$\frac{3 x - 1 - 2 - 2 (- 5 x)}{1 - 5 x} > 0$

Step 2.2.4.3

Multiply $- 5$ by $- 2$ .

$\frac{3 x - 1 - 2 + 10 x}{1 - 5 x} > 0$

Step 2.2.4.4

Add $3 x$ and $10 x$ .

$\frac{13 x - 1 - 2}{1 - 5 x} > 0$

Step 2.2.4.5

Subtract $2$ from $- 1$ .

$\frac{13 x - 3}{1 - 5 x} > 0$

Step 2.3

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

$13 x - 3 = 0$

$1 - 5 x = 0$

Step 2.4

Add $3$ to both sides of the equation.

$13 x = 3$

Step 2.5

Divide each term in $13 x = 3$ by $13$ and simplify.

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Step 2.5.1

Divide each term in $13 x = 3$ by $13$ .

$\frac{13 x}{13} = \frac{3}{13}$

Step 2.5.2

Simplify the left side.

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Step 2.5.2.1

Cancel the common factor of $13$ .

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Step 2.5.2.1.1

Cancel the common factor.

$\frac{13 x}{13} = \frac{3}{13}$

Step 2.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{3}{13}$

Step 2.6

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

Step 2.7

Divide each term in $- 5 x = - 1$ by $- 5$ and simplify.

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Step 2.7.1

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 2.7.2

Simplify the left side.

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Step 2.7.2.1

Cancel the common factor of $- 5$ .

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Step 2.7.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 2.7.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Step 2.7.3

Simplify the right side.

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Step 2.7.3.1

Dividing two negative values results in a positive value.

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

Step 2.8

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

$x = \frac{3}{13}$

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

Step 2.9

Consolidate the solutions.

$x = \frac{3}{13}, \frac{1}{5}$

Step 2.10

Find the domain of $\frac{13 x - 3}{1 - 5 x}$ .

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Step 2.10.1

Set the denominator in $\frac{13 x - 3}{1 - 5 x}$ equal to $0$ to find where the expression is undefined.

$1 - 5 x = 0$

Step 2.10.2

Solve for $x$ .

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Step 2.10.2.1

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

Step 2.10.2.2

Divide each term in $- 5 x = - 1$ by $- 5$ and simplify.

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Step 2.10.2.2.1

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 2.10.2.2.2

Simplify the left side.

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Step 2.10.2.2.2.1

Cancel the common factor of $- 5$ .

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Step 2.10.2.2.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 2.10.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Step 2.10.2.2.3

Simplify the right side.

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Step 2.10.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 2.10.3

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

$(- \infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$

Step 2.11

Use each root to create test intervals.

$x < \frac{1}{5}$

$\frac{1}{5} < x < \frac{3}{13}$

$x > \frac{3}{13}$

Step 2.12

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 2.12.1

Test a value on the interval $x < \frac{1}{5}$ to see if it makes the inequality true.

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Step 2.12.1.1

Choose a value on the interval $x < \frac{1}{5}$ and see if this value makes the original inequality true.

$x = 0$

Step 2.12.1.2

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

$\frac{3 (0) - 1}{1 - 5 \cdot 0} > 2$

Step 2.12.1.3

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

False

Step 2.12.2

Test a value on the interval $\frac{1}{5} < x < \frac{3}{13}$ to see if it makes the inequality true.

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Step 2.12.2.1

Choose a value on the interval $\frac{1}{5} < x < \frac{3}{13}$ and see if this value makes the original inequality true.

$x = 0.22$

Step 2.12.2.2

Replace $x$ with $0.22$ in the original inequality.

$\frac{3 (0.22) - 1}{1 - 5 \cdot 0.22} > 2$

Step 2.12.2.3

The left side $3.4$ is greater than the right side $2$ , which means that the given statement is always true.

True

Step 2.12.3

Test a value on the interval $x > \frac{3}{13}$ to see if it makes the inequality true.

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Step 2.12.3.1

Choose a value on the interval $x > \frac{3}{13}$ and see if this value makes the original inequality true.

$x = 3$

Step 2.12.3.2

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

$\frac{3 (3) - 1}{1 - 5 \cdot 3} > 2$

Step 2.12.3.3

The left side $- 0.\overline{571428}$ is not greater than the right side $2$ , which means that the given statement is false.

False

Step 2.12.4

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

$x < \frac{1}{5}$ False

$\frac{1}{5} < x < \frac{3}{13}$ True

$x > \frac{3}{13}$ False

$x < \frac{1}{5}$ False

$\frac{1}{5} < x < \frac{3}{13}$ True

$x > \frac{3}{13}$ False

Step 2.13

The solution consists of all of the true intervals.

$\frac{1}{5} < x < \frac{3}{13}$

Step 3

Solve $- \frac{3 x - 1}{1 - 5 x} > 2$ for $x$ .

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Step 3.1

Subtract $2$ from both sides of the inequality.

$- \frac{3 x - 1}{1 - 5 x} - 2 > 0$

Step 3.2

Simplify $- \frac{3 x - 1}{1 - 5 x} - 2$ .

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Step 3.2.1

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{1 - 5 x}{1 - 5 x}$ .

$- \frac{3 x - 1}{1 - 5 x} - 2 \cdot \frac{1 - 5 x}{1 - 5 x} > 0$

Step 3.2.2

Combine $- 2$ and $\frac{1 - 5 x}{1 - 5 x}$ .

$- \frac{3 x - 1}{1 - 5 x} + \frac{- 2 (1 - 5 x)}{1 - 5 x} > 0$

Step 3.2.3

Combine the numerators over the common denominator.

$\frac{- (3 x - 1) - 2 (1 - 5 x)}{1 - 5 x} > 0$

Step 3.2.4

Simplify the numerator.

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Step 3.2.4.1

Factor $- 1$ out of $- (3 x - 1) - 2 (1 - 5 x)$ .

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Step 3.2.4.1.1

Reorder $1$ and $- 5 x$ .

$\frac{- (3 x - 1) - 2 (- 5 x + 1)}{1 - 5 x} > 0$

Step 3.2.4.1.2

Factor $- 1$ out of $- 2 (- 5 x + 1)$ .

$\frac{- (3 x - 1) - (2 (- 5 x + 1))}{1 - 5 x} > 0$

Step 3.2.4.1.3

Factor $- 1$ out of $- (3 x - 1) - (2 (- 5 x + 1))$ .

$\frac{- (3 x - 1 + 2 (- 5 x + 1))}{1 - 5 x} > 0$

Step 3.2.4.2

Apply the distributive property.

$\frac{- (3 x - 1 + 2 (- 5 x) + 2 \cdot 1)}{1 - 5 x} > 0$

Step 3.2.4.3

Multiply $- 5$ by $2$ .

$\frac{- (3 x - 1 - 10 x + 2 \cdot 1)}{1 - 5 x} > 0$

Step 3.2.4.4

Multiply $2$ by $1$ .

$\frac{- (3 x - 1 - 10 x + 2)}{1 - 5 x} > 0$

Step 3.2.4.5

Subtract $10 x$ from $3 x$ .

$\frac{- (- 7 x - 1 + 2)}{1 - 5 x} > 0$

Step 3.2.4.6

Add $- 1$ and $2$ .

$\frac{- (- 7 x + 1)}{1 - 5 x} > 0$

Step 3.2.5

Move the negative in front of the fraction.

$- \frac{- 7 x + 1}{1 - 5 x} > 0$

Step 3.2.6

Factor $- 1$ out of $- 7 x$ .

$- \frac{- (7 x) + 1}{1 - 5 x} > 0$

Step 3.2.7

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

$- \frac{- (7 x) - 1 (- 1)}{1 - 5 x} > 0$

Step 3.2.8

Factor $- 1$ out of $- (7 x) - 1 (- 1)$ .

$- \frac{- (7 x - 1)}{1 - 5 x} > 0$

Step 3.2.9

Rewrite $- (7 x - 1)$ as $- 1 (7 x - 1)$ .

$- \frac{- 1 (7 x - 1)}{1 - 5 x} > 0$

Step 3.2.10

Move the negative in front of the fraction.

$- - \frac{7 x - 1}{1 - 5 x} > 0$

Step 3.2.11

Multiply $- 1$ by $- 1$ .

$1 \frac{7 x - 1}{1 - 5 x} > 0$

Step 3.2.12

Multiply $\frac{7 x - 1}{1 - 5 x}$ by $1$ .

$\frac{7 x - 1}{1 - 5 x} > 0$

Step 3.3

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

$7 x - 1 = 0$

$1 - 5 x = 0$

Step 3.4

Add $1$ to both sides of the equation.

$7 x = 1$

Step 3.5

Divide each term in $7 x = 1$ by $7$ and simplify.

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Step 3.5.1

Divide each term in $7 x = 1$ by $7$ .

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

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Cancel the common factor of $7$ .

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Step 3.5.2.1.1

Cancel the common factor.

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

Step 3.5.2.1.2

Divide $x$ by $1$ .

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

Step 3.6

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

Step 3.7

Divide each term in $- 5 x = - 1$ by $- 5$ and simplify.

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Step 3.7.1

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 3.7.2

Simplify the left side.

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Step 3.7.2.1

Cancel the common factor of $- 5$ .

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Step 3.7.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 3.7.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Step 3.7.3

Simplify the right side.

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Step 3.7.3.1

Dividing two negative values results in a positive value.

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

Step 3.8

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

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

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

Step 3.9

Consolidate the solutions.

$x = \frac{1}{7}, \frac{1}{5}$

Step 3.10

Find the domain of $\frac{7 x - 1}{1 - 5 x}$ .

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Step 3.10.1

Set the denominator in $\frac{7 x - 1}{1 - 5 x}$ equal to $0$ to find where the expression is undefined.

$1 - 5 x = 0$

Step 3.10.2

Solve for $x$ .

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Step 3.10.2.1

Subtract $1$ from both sides of the equation.

$- 5 x = - 1$

Step 3.10.2.2

Divide each term in $- 5 x = - 1$ by $- 5$ and simplify.

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Step 3.10.2.2.1

Divide each term in $- 5 x = - 1$ by $- 5$ .

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 3.10.2.2.2

Simplify the left side.

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Step 3.10.2.2.2.1

Cancel the common factor of $- 5$ .

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Step 3.10.2.2.2.1.1

Cancel the common factor.

$\frac{- 5 x}{- 5} = \frac{- 1}{- 5}$

Step 3.10.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 5}$

Step 3.10.2.2.3

Simplify the right side.

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Step 3.10.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 3.10.3

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

$(- \infty, \frac{1}{5}) \cup (\frac{1}{5}, \infty)$

Step 3.11

Use each root to create test intervals.

$x < \frac{1}{7}$

$\frac{1}{7} < x < \frac{1}{5}$

$x > \frac{1}{5}$

Step 3.12

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 3.12.1

Test a value on the interval $x < \frac{1}{7}$ to see if it makes the inequality true.

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Step 3.12.1.1

Choose a value on the interval $x < \frac{1}{7}$ and see if this value makes the original inequality true.

$x = 0$

Step 3.12.1.2

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

$- \frac{3 (0) - 1}{1 - 5 \cdot 0} > 2$

Step 3.12.1.3

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

False

Step 3.12.2

Test a value on the interval $\frac{1}{7} < x < \frac{1}{5}$ to see if it makes the inequality true.

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Step 3.12.2.1

Choose a value on the interval $\frac{1}{7} < x < \frac{1}{5}$ and see if this value makes the original inequality true.

$x = 0.17$

Step 3.12.2.2

Replace $x$ with $0.17$ in the original inequality.

$- \frac{3 (0.17) - 1}{1 - 5 \cdot 0.17} > 2$

Step 3.12.2.3

The left side $3.2\overline{6}$ is greater than the right side $2$ , which means that the given statement is always true.

True

Step 3.12.3

Test a value on the interval $x > \frac{1}{5}$ to see if it makes the inequality true.

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Step 3.12.3.1

Choose a value on the interval $x > \frac{1}{5}$ and see if this value makes the original inequality true.

$x = 3$

Step 3.12.3.2

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

$- \frac{3 (3) - 1}{1 - 5 \cdot 3} > 2$

Step 3.12.3.3

The left side $0.\overline{571428}$ is not greater than the right side $2$ , which means that the given statement is false.

False

Step 3.12.4

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

$x < \frac{1}{7}$ False

$\frac{1}{7} < x < \frac{1}{5}$ True

$x > \frac{1}{5}$ False

$x < \frac{1}{7}$ False

$\frac{1}{7} < x < \frac{1}{5}$ True

$x > \frac{1}{5}$ False

Step 3.13

The solution consists of all of the true intervals.

$\frac{1}{7} < x < \frac{1}{5}$

Step 4

Find the union of the solutions.

$\frac{1}{7} < x < \frac{1}{5}$ or $\frac{1}{5} < x < \frac{3}{13}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$\frac{1}{7} < x < \frac{1}{5} or \frac{1}{5} < x < \frac{3}{13}$

Interval Notation:

$(\frac{1}{7}, \frac{1}{5}) \cup (\frac{1}{5}, \frac{3}{13})$

Step 6