Convert to Set Notation x/4<=9/x

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

Step 1

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

Multiply both sides by $4$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

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

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

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

Multiply $9$ by $4$ .

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

Solve for $x$ .

Multiply both sides by $x$ .

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

Simplify.

Simplify the left side.

Multiply $x$ by $x$ .

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

Simplify the right side.

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

$x^{2} = 36$

Solve for $x$ .

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

$x = \pm \sqrt{36}$

Simplify $\pm \sqrt{36}$ .

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

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

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

$x = \pm 6$

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

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

$x = 6$

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

$x = - 6$

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

$x = 6, - 6$

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

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

$x = 0$

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

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

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 0$

$0 < x < 6$

$x > 6$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = - 8$

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

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

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

True

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

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

$x = - 3$

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

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

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

False

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

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

$x = 3$

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

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

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

True

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

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

$x = 8$

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

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

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

False

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

The solution consists of all of the true intervals.

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

Step 2

Use the inequality $x \leq - 6 or 0 < x \leq 6$ to build the set notation.

${x | x \leq - 6 or 0 < x \leq 6}$

Step 3

Convert to Set Notation x/4<=9/x

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