Algebra Examples

$\frac{x}{4} \leq \frac{9}{x}$

Step 1

Multiply both sides by $4$ .

$\frac{x}{4} \cdot 4 \leq \frac{9}{x} \cdot 4$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Cancel the common factor of $4$ .

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Step 2.1.1.1

Cancel the common factor.

$\frac{x}{4} \cdot 4 \leq \frac{9}{x} \cdot 4$

Step 2.1.1.2

Rewrite the expression.

$x \leq \frac{9}{x} \cdot 4$

Step 2.2

Simplify the right side.

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Step 2.2.1

Multiply $\frac{9}{x} \cdot 4$ .

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Step 2.2.1.1

Combine $\frac{9}{x}$ and $4$ .

$x \leq \frac{9 \cdot 4}{x}$

Step 2.2.1.2

Multiply $9$ by $4$ .

$x \leq \frac{36}{x}$

Step 3

Solve for $x$ .

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Step 3.1

Multiply both sides by $x$ .

$x \cdot x = \frac{36}{x} x$

Step 3.2

Simplify.

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Step 3.2.1

Simplify the left side.

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Step 3.2.1.1

Multiply $x$ by $x$ .

$x^{2} = \frac{36}{x} x$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

Cancel the common factor of $x$ .

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Step 3.2.2.1.1

Cancel the common factor.

$x^{2} = \frac{36}{x} x$

Step 3.2.2.1.2

Rewrite the expression.

$x^{2} = 36$

Step 3.3

Solve for $x$ .

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Step 3.3.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{36}$

Step 3.3.2

Simplify $\pm \sqrt{36}$ .

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Step 3.3.2.1

Rewrite $36$ as $6^{2}$ .

$x = \pm \sqrt{6^{2}}$

Step 3.3.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 6$

Step 3.3.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.3.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 6$

Step 3.3.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 6$

Step 3.3.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 6, - 6$

Step 4

Find the domain of $\frac{x}{4} - \frac{9}{x}$ .

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Step 4.1

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

$x = 0$

Step 4.2

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

$(- \infty, 0) \cup (0, \infty)$

Step 5

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 0$

$0 < x < 6$

$x > 6$

Step 6

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 6.1

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

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Step 6.1.1

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

$x = - 8$

Step 6.1.2

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

$\frac{- 8}{4} \leq \frac{9}{- 8}$

Step 6.1.3

The left side $- 2$ is less than the right side $- 1.125$ , which means that the given statement is always true.

True

Step 6.2

Test a value on the interval $- 6 < x < 0$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $- 6 < x < 0$ and see if this value makes the original inequality true.

$x = - 3$

Step 6.2.2

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

$\frac{- 3}{4} \leq \frac{9}{- 3}$

Step 6.2.3

The left side $- 0.75$ is greater than the right side $- 3$ , which means that the given statement is false.

False

Step 6.3

Test a value on the interval $0 < x < 6$ to see if it makes the inequality true.

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Step 6.3.1

Choose a value on the interval $0 < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Step 6.3.2

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

$\frac{3}{4} \leq \frac{9}{3}$

Step 6.3.3

The left side $0.75$ is less than the right side $3$ , which means that the given statement is always true.

True

Step 6.4

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

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Step 6.4.1

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

$x = 8$

Step 6.4.2

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

$\frac{8}{4} \leq \frac{9}{8}$

Step 6.4.3

The left side $2$ is greater than the right side $1.125$ , which means that the given statement is false.

False

Step 6.5

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

$x < - 6$ True

$- 6 < x < 0$ False

$0 < x < 6$ True

$x > 6$ False

$x < - 6$ True

$- 6 < x < 0$ False

$0 < x < 6$ True

$x > 6$ False

Step 7

The solution consists of all of the true intervals.

$x \leq - 6$ or $0 < x \leq 6$

Step 8

Convert the inequality to interval notation.

$(- \infty, - 6] \cup (0, 6]$

Step 9