Algebra Examples

$\frac{5}{x} + \frac{6}{5 x} > \frac{2}{3}$

Step 1

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

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Step 1.1

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

$\frac{5}{x} \cdot \frac{5}{5} + \frac{6}{5 x} > \frac{2}{3}$

Step 1.2

Write each expression with a common denominator of $x \cdot 5$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.2.1

Multiply $\frac{5}{x}$ by $\frac{5}{5}$ .

$\frac{5 \cdot 5}{x \cdot 5} + \frac{6}{5 x} > \frac{2}{3}$

Step 1.2.2

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

$\frac{5 \cdot 5}{5 x} + \frac{6}{5 x} > \frac{2}{3}$

Step 1.3

Combine the numerators over the common denominator.

$\frac{5 \cdot 5 + 6}{5 x} > \frac{2}{3}$

Step 1.4

Simplify the numerator.

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Step 1.4.1

Multiply $5$ by $5$ .

$\frac{25 + 6}{5 x} > \frac{2}{3}$

Step 1.4.2

Add $25$ and $6$ .

$\frac{31}{5 x} > \frac{2}{3}$

Step 2

Multiply both sides by $5 x$ .

$\frac{31}{5 x} (5 x) = \frac{2}{3} (5 x)$

Step 3

Simplify.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Simplify $\frac{31}{5 x} (5 x)$ .

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Step 3.1.1.1

Rewrite using the commutative property of multiplication.

$5 \frac{31}{5 x} x = \frac{2}{3} (5 x)$

Step 3.1.1.2

Cancel the common factor of $5$ .

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Step 3.1.1.2.1

Factor $5$ out of $5 x$ .

$5 \frac{31}{5 (x)} x = \frac{2}{3} (5 x)$

Step 3.1.1.2.2

Cancel the common factor.

$5 \frac{31}{5 x} x = \frac{2}{3} (5 x)$

Step 3.1.1.2.3

Rewrite the expression.

$\frac{31}{x} x = \frac{2}{3} (5 x)$

Step 3.1.1.3

Cancel the common factor of $x$ .

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Step 3.1.1.3.1

Cancel the common factor.

$\frac{31}{x} x = \frac{2}{3} (5 x)$

Step 3.1.1.3.2

Rewrite the expression.

$31 = \frac{2}{3} (5 x)$

Step 3.2

Simplify the right side.

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Step 3.2.1

Multiply $\frac{2}{3} (5 x)$ .

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Step 3.2.1.1

Combine $5$ and $\frac{2}{3}$ .

$31 = \frac{5 \cdot 2}{3} x$

Step 3.2.1.2

Multiply $5$ by $2$ .

$31 = \frac{10}{3} x$

Step 3.2.1.3

Combine $\frac{10}{3}$ and $x$ .

$31 = \frac{10 x}{3}$

Step 4

Solve for $x$ .

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Step 4.1

Rewrite the equation as $\frac{10 x}{3} = 31$ .

$\frac{10 x}{3} = 31$

Step 4.2

Multiply both sides of the equation by $\frac{3}{10}$ .

$\frac{3}{10} \cdot \frac{10 x}{3} = \frac{3}{10} \cdot 31$

Step 4.3

Simplify both sides of the equation.

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Step 4.3.1

Simplify the left side.

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Step 4.3.1.1

Simplify $\frac{3}{10} \cdot \frac{10 x}{3}$ .

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Step 4.3.1.1.1

Cancel the common factor of $3$ .

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Step 4.3.1.1.1.1

Cancel the common factor.

$\frac{3}{10} \cdot \frac{10 x}{3} = \frac{3}{10} \cdot 31$

Step 4.3.1.1.1.2

Rewrite the expression.

$\frac{1}{10} (10 x) = \frac{3}{10} \cdot 31$

Step 4.3.1.1.2

Cancel the common factor of $10$ .

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Step 4.3.1.1.2.1

Factor $10$ out of $10 x$ .

$\frac{1}{10} (10 (x)) = \frac{3}{10} \cdot 31$

Step 4.3.1.1.2.2

Cancel the common factor.

$\frac{1}{10} (10 x) = \frac{3}{10} \cdot 31$

Step 4.3.1.1.2.3

Rewrite the expression.

$x = \frac{3}{10} \cdot 31$

Step 4.3.2

Simplify the right side.

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Step 4.3.2.1

Multiply $\frac{3}{10} \cdot 31$ .

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Step 4.3.2.1.1

Combine $\frac{3}{10}$ and $31$ .

$x = \frac{3 \cdot 31}{10}$

Step 4.3.2.1.2

Multiply $3$ by $31$ .

$x = \frac{93}{10}$

Step 5

Find the domain of $\frac{31}{5 x} - \frac{2}{3}$ .

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Step 5.1

Set the denominator in $\frac{31}{5 x}$ equal to $0$ to find where the expression is undefined.

$5 x = 0$

Step 5.2

Divide each term in $5 x = 0$ by $5$ and simplify.

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Step 5.2.1

Divide each term in $5 x = 0$ by $5$ .

$\frac{5 x}{5} = \frac{0}{5}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $5$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{0}{5}$

Step 5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{5}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Divide $0$ by $5$ .

$x = 0$

Step 5.3

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

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

Step 6

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{93}{10}$

$x > \frac{93}{10}$

Step 7

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 7.1

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

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Step 7.1.1

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

$x = - 2$

Step 7.1.2

Replace $x$ with $- 2$ in the original inequality.

$\frac{5}{- 2} + \frac{6}{5 (- 2)} > \frac{2}{3}$

Step 7.1.3

The left side $- 3.1$ is not greater than the right side $0.\overline{6}$ , which means that the given statement is false.

False

Step 7.2

Test a value on the interval $0 < x < \frac{93}{10}$ to see if it makes the inequality true.

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Step 7.2.1

Choose a value on the interval $0 < x < \frac{93}{10}$ and see if this value makes the original inequality true.

$x = 5$

Step 7.2.2

Replace $x$ with $5$ in the original inequality.

$\frac{5}{5} + \frac{6}{5 (5)} > \frac{2}{3}$

Step 7.2.3

The left side $1.24$ is greater than the right side $0.\overline{6}$ , which means that the given statement is always true.

True

Step 7.3

Test a value on the interval $x > \frac{93}{10}$ to see if it makes the inequality true.

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Step 7.3.1

Choose a value on the interval $x > \frac{93}{10}$ and see if this value makes the original inequality true.

$x = 12$

Step 7.3.2

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

$\frac{5}{12} + \frac{6}{5 (12)} > \frac{2}{3}$

Step 7.3.3

The left side $0.51\overline{6}$ is not greater than the right side $0.\overline{6}$ , which means that the given statement is false.

False

Step 7.4

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

$x < 0$ False

$0 < x < \frac{93}{10}$ True

$x > \frac{93}{10}$ False

$x < 0$ False

$0 < x < \frac{93}{10}$ True

$x > \frac{93}{10}$ False

Step 8

The solution consists of all of the true intervals.

$0 < x < \frac{93}{10}$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$0 < x < \frac{93}{10}$

Interval Notation:

$(0, \frac{93}{10})$

Step 10