Pre-Algebra Examples

$\frac{x - 13}{x - 20} > 3$

Step 1

Subtract $3$ from both sides of the inequality.

$\frac{x - 13}{x - 20} - 3 > 0$

Step 2

Simplify $\frac{x - 13}{x - 20} - 3$ .

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Step 2.1

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{x - 20}{x - 20}$ .

$\frac{x - 13}{x - 20} - 3 \cdot \frac{x - 20}{x - 20} > 0$

Step 2.2

Combine $- 3$ and $\frac{x - 20}{x - 20}$ .

$\frac{x - 13}{x - 20} + \frac{- 3 (x - 20)}{x - 20} > 0$

Step 2.3

Combine the numerators over the common denominator.

$\frac{x - 13 - 3 (x - 20)}{x - 20} > 0$

Step 2.4

Simplify the numerator.

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Step 2.4.1

Apply the distributive property.

$\frac{x - 13 - 3 x - 3 \cdot - 20}{x - 20} > 0$

Step 2.4.2

Multiply $- 3$ by $- 20$ .

$\frac{x - 13 - 3 x + 60}{x - 20} > 0$

Step 2.4.3

Subtract $3 x$ from $x$ .

$\frac{- 2 x - 13 + 60}{x - 20} > 0$

Step 2.4.4

Add $- 13$ and $60$ .

$\frac{- 2 x + 47}{x - 20} > 0$

Step 2.5

Factor $- 1$ out of $- 2 x$ .

$\frac{- (2 x) + 47}{x - 20} > 0$

Step 2.6

Rewrite $47$ as $- 1 (- 47)$ .

$\frac{- (2 x) - 1 (- 47)}{x - 20} > 0$

Step 2.7

Factor $- 1$ out of $- (2 x) - 1 (- 47)$ .

$\frac{- (2 x - 47)}{x - 20} > 0$

Step 2.8

Rewrite $- (2 x - 47)$ as $- 1 (2 x - 47)$ .

$\frac{- 1 (2 x - 47)}{x - 20} > 0$

Step 2.9

Move the negative in front of the fraction.

$- \frac{2 x - 47}{x - 20} > 0$

Step 3

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

$2 x - 47 = 0$

$x - 20 = 0$

Step 4

Add $47$ to both sides of the equation.

$2 x = 47$

Step 5

Divide each term in $2 x = 47$ by $2$ and simplify.

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Step 5.1

Divide each term in $2 x = 47$ by $2$ .

$\frac{2 x}{2} = \frac{47}{2}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $2$ .

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Step 5.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{47}{2}$

Step 5.2.1.2

Divide $x$ by $1$ .

$x = \frac{47}{2}$

Step 6

Add $20$ to both sides of the equation.

$x = 20$

Step 7

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

$x = \frac{47}{2}$

$x = 20$

Step 8

Consolidate the solutions.

$x = \frac{47}{2}, 20$

Step 9

Find the domain of $- \frac{2 x - 47}{x - 20}$ .

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Step 9.1

Set the denominator in $\frac{2 x - 47}{x - 20}$ equal to $0$ to find where the expression is undefined.

$x - 20 = 0$

Step 9.2

Add $20$ to both sides of the equation.

$x = 20$

Step 9.3

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

$(- \infty, 20) \cup (20, \infty)$

Step 10

Use each root to create test intervals.

$x < 20$

$20 < x < \frac{47}{2}$

$x > \frac{47}{2}$

Step 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 11.1

Test a value on the interval $x < 20$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $x < 20$ and see if this value makes the original inequality true.

$x = 0$

Step 11.1.2

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

$\frac{(0) - 13}{(0) - 20} > 3$

Step 11.1.3

The left side $0.65$ is not greater than the right side $3$ , which means that the given statement is false.

False

Step 11.2

Test a value on the interval $20 < x < \frac{47}{2}$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $20 < x < \frac{47}{2}$ and see if this value makes the original inequality true.

$x = 22$

Step 11.2.2

Replace $x$ with $22$ in the original inequality.

$\frac{(22) - 13}{(22) - 20} > 3$

Step 11.2.3

The left side $4.5$ is greater than the right side $3$ , which means that the given statement is always true.

True

Step 11.3

Test a value on the interval $x > \frac{47}{2}$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $x > \frac{47}{2}$ and see if this value makes the original inequality true.

$x = 26$

Step 11.3.2

Replace $x$ with $26$ in the original inequality.

$\frac{(26) - 13}{(26) - 20} > 3$

Step 11.3.3

The left side $2.1\overline{6}$ is not greater than the right side $3$ , which means that the given statement is false.

False

Step 11.4

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

$x < 20$ False

$20 < x < \frac{47}{2}$ True

$x > \frac{47}{2}$ False

$x < 20$ False

$20 < x < \frac{47}{2}$ True

$x > \frac{47}{2}$ False

Step 12

The solution consists of all of the true intervals.

$20 < x < \frac{47}{2}$

Step 13