Precalculus Examples

$6 x < \frac{36}{x + 5}$

Step 1

Subtract $\frac{36}{x + 5}$ from both sides of the inequality.

$6 x - \frac{36}{x + 5} < 0$

Step 2

Simplify $6 x - \frac{36}{x + 5}$ .

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Step 2.1

To write $6 x$ as a fraction with a common denominator, multiply by $\frac{x + 5}{x + 5}$ .

$\frac{6 x (x + 5)}{x + 5} - \frac{36}{x + 5} < 0$

Step 2.2

Combine the numerators over the common denominator.

$\frac{6 x (x + 5) - 36}{x + 5} < 0$

Step 2.3

Simplify the numerator.

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Step 2.3.1

Factor $6$ out of $6 x (x + 5) - 36$ .

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Step 2.3.1.1

Factor $6$ out of $6 x (x + 5)$ .

$\frac{6 (x (x + 5)) - 36}{x + 5} < 0$

Step 2.3.1.2

Factor $6$ out of $- 36$ .

$\frac{6 (x (x + 5)) + 6 \cdot - 6}{x + 5} < 0$

Step 2.3.1.3

Factor $6$ out of $6 (x (x + 5)) + 6 \cdot - 6$ .

$\frac{6 (x (x + 5) - 6)}{x + 5} < 0$

Step 2.3.2

Apply the distributive property.

$\frac{6 (x \cdot x + x \cdot 5 - 6)}{x + 5} < 0$

Step 2.3.3

Multiply $x$ by $x$ .

$\frac{6 (x^{2} + x \cdot 5 - 6)}{x + 5} < 0$

Step 2.3.4

Move $5$ to the left of $x$ .

$\frac{6 (x^{2} + 5 \cdot x - 6)}{x + 5} < 0$

Step 2.3.5

Factor $x^{2} + 5 x - 6$ using the AC method.

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Step 2.3.5.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $5$ .

$- 1, 6$

Step 2.3.5.2

Write the factored form using these integers.

$\frac{6 ((x - 1) (x + 6))}{x + 5} < 0$

$\frac{6 (x - 1) (x + 6)}{x + 5} < 0$

Step 3

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

$x - 1 = 0$

$x + 6 = 0$

$x + 5 = 0$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 6

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 7

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

$x = 1$

$x = - 6$

$x = - 5$

Step 8

Consolidate the solutions.

$x = 1, - 6, - 5$

Step 9

Find the domain of $\frac{6 (x - 1) (x + 6)}{x + 5}$ .

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Step 9.1

Set the denominator in $\frac{6 (x - 1) (x + 6)}{x + 5}$ equal to $0$ to find where the expression is undefined.

$x + 5 = 0$

Step 9.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 9.3

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

$(- \infty, - 5) \cup (- 5, \infty)$

Step 10

Use each root to create test intervals.

$x < - 6$

$- 6 < x < - 5$

$- 5 < x < 1$

$x > 1$

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 $x < - 6$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $x < - 6$ and see if this value makes the original inequality true.

$x = - 8$

Step 11.1.2

Replace $x$ with $- 8$ in the original inequality.

$6 (- 8) < \frac{36}{(- 8) + 5}$

Step 11.1.3

The left side $- 48$ is less than the right side $- 12$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $- 6 < x < - 5$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $- 6 < x < - 5$ and see if this value makes the original inequality true.

$x = - 5.5$

Step 11.2.2

Replace $x$ with $- 5.5$ in the original inequality.

$6 (- 5.5) < \frac{36}{(- 5.5) + 5}$

Step 11.2.3

The left side $- 33$ is not less than the right side $- 72$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $- 5 < x < 1$ to see if it makes the inequality true.

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Step 11.3.1

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

$x = 0$

Step 11.3.2

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

$6 (0) < \frac{36}{(0) + 5}$

Step 11.3.3

The left side $0$ is less than the right side $7.2$ , which means that the given statement is always true.

True

Step 11.4

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 11.4.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 11.4.2

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

$6 (4) < \frac{36}{(4) + 5}$

Step 11.4.3

The left side $24$ is not less than the right side $4$ , which means that the given statement is false.

False

Step 11.5

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

$x < - 6$ True

$- 6 < x < - 5$ False

$- 5 < x < 1$ True

$x > 1$ False

$x < - 6$ True

$- 6 < x < - 5$ False

$- 5 < x < 1$ True

$x > 1$ False

Step 12

The solution consists of all of the true intervals.

$x < - 6$ or $- 5 < x < 1$

Step 13

Convert the inequality to interval notation.

$(- \infty, - 6) \cup (- 5, 1)$

Step 14