Algebra Examples

$\frac{x + 4}{x + 9} < 2$

Step 1

Subtract $2$ from both sides of the inequality.

$\frac{x + 4}{x + 9} - 2 < 0$

Step 2

Simplify $\frac{x + 4}{x + 9} - 2$ .

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Step 2.1

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{x + 9}{x + 9}$ .

$\frac{x + 4}{x + 9} - 2 \cdot \frac{x + 9}{x + 9} < 0$

Step 2.2

Combine $- 2$ and $\frac{x + 9}{x + 9}$ .

$\frac{x + 4}{x + 9} + \frac{- 2 (x + 9)}{x + 9} < 0$

Step 2.3

Combine the numerators over the common denominator.

$\frac{x + 4 - 2 (x + 9)}{x + 9} < 0$

Step 2.4

Simplify the numerator.

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Step 2.4.1

Apply the distributive property.

$\frac{x + 4 - 2 x - 2 \cdot 9}{x + 9} < 0$

Step 2.4.2

Multiply $- 2$ by $9$ .

$\frac{x + 4 - 2 x - 18}{x + 9} < 0$

Step 2.4.3

Subtract $2 x$ from $x$ .

$\frac{- x + 4 - 18}{x + 9} < 0$

Step 2.4.4

Subtract $18$ from $4$ .

$\frac{- x - 14}{x + 9} < 0$

Step 2.5

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

$\frac{- (x) - 14}{x + 9} < 0$

Step 2.6

Rewrite $- 14$ as $- 1 (14)$ .

$\frac{- (x) - 1 (14)}{x + 9} < 0$

Step 2.7

Factor $- 1$ out of $- (x) - 1 (14)$ .

$\frac{- (x + 14)}{x + 9} < 0$

Step 2.8

Rewrite $- (x + 14)$ as $- 1 (x + 14)$ .

$\frac{- 1 (x + 14)}{x + 9} < 0$

Step 2.9

Move the negative in front of the fraction.

$- \frac{x + 14}{x + 9} < 0$

Step 3

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

$x + 14 = 0$

$x + 9 = 0$

Step 4

Subtract $14$ from both sides of the equation.

$x = - 14$

Step 5

Subtract $9$ from both sides of the equation.

$x = - 9$

Step 6

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

$x = - 14$

$x = - 9$

Step 7

Consolidate the solutions.

$x = - 14, - 9$

Step 8

Find the domain of $- \frac{x + 14}{x + 9}$ .

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Step 8.1

Set the denominator in $\frac{x + 14}{x + 9}$ equal to $0$ to find where the expression is undefined.

$x + 9 = 0$

Step 8.2

Subtract $9$ from both sides of the equation.

$x = - 9$

Step 8.3

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

$(- \infty, - 9) \cup (- 9, \infty)$

Step 9

Use each root to create test intervals.

$x < - 14$

$- 14 < x < - 9$

$x > - 9$

Step 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 10.1

Test a value on the interval $x < - 14$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $x < - 14$ and see if this value makes the original inequality true.

$x = - 16$

Step 10.1.2

Replace $x$ with $- 16$ in the original inequality.

$\frac{(- 16) + 4}{(- 16) + 9} < 2$

Step 10.1.3

The left side $1.\overline{714285}$ is less than the right side $2$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $- 14 < x < - 9$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 14 < x < - 9$ and see if this value makes the original inequality true.

$x = - 12$

Step 10.2.2

Replace $x$ with $- 12$ in the original inequality.

$\frac{(- 12) + 4}{(- 12) + 9} < 2$

Step 10.2.3

The left side $2.\overline{6}$ is not less than the right side $2$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $x > - 9$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $x > - 9$ and see if this value makes the original inequality true.

$x = 0$

Step 10.3.2

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

$\frac{(0) + 4}{(0) + 9} < 2$

Step 10.3.3

The left side $0.\overline{4}$ is less than the right side $2$ , which means that the given statement is always true.

True

Step 10.4

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

$x < - 14$ True

$- 14 < x < - 9$ False

$x > - 9$ True

$x < - 14$ True

$- 14 < x < - 9$ False

$x > - 9$ True

Step 11

The solution consists of all of the true intervals.

$x < - 14$ or $x > - 9$

Step 12

Convert the inequality to interval notation.

$(- \infty, - 14) \cup (- 9, \infty)$

Step 13