Algebra Examples

$5 \leq \frac{- 2}{- 5 x} + \frac{4}{2}$

Step 1

Rewrite so $x$ is on the left side of the inequality.

$\frac{- 2}{- 5 x} + \frac{4}{2} \geq 5$

Step 2

Simplify each term.

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Step 2.1

Dividing two negative values results in a positive value.

$\frac{2}{5 x} + \frac{4}{2} \geq 5$

Step 2.2

Divide $4$ by $2$ .

$\frac{2}{5 x} + 2 \geq 5$

Step 3

Move all terms not containing $x$ to the right side of the inequality.

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Step 3.1

Subtract $2$ from both sides of the inequality.

$\frac{2}{5 x} \geq 5 - 2$

Step 3.2

Subtract $2$ from $5$ .

$\frac{2}{5 x} \geq 3$

Step 4

Multiply both sides by $5 x$ .

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

Step 5

Simplify.

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Step 5.1

Simplify the left side.

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Step 5.1.1

Simplify $\frac{2}{5 x} (5 x)$ .

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Step 5.1.1.1

Rewrite using the commutative property of multiplication.

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

Step 5.1.1.2

Cancel the common factor of $5$ .

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Step 5.1.1.2.1

Factor $5$ out of $5 x$ .

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

Step 5.1.1.2.2

Cancel the common factor.

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

Step 5.1.1.2.3

Rewrite the expression.

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

Step 5.1.1.3

Cancel the common factor of $x$ .

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Step 5.1.1.3.1

Cancel the common factor.

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

Step 5.1.1.3.2

Rewrite the expression.

$2 = 3 (5 x)$

Step 5.2

Simplify the right side.

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Step 5.2.1

Multiply $5$ by $3$ .

$2 = 15 x$

Step 6

Solve for $x$ .

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Step 6.1

Rewrite the equation as $15 x = 2$ .

$15 x = 2$

Step 6.2

Divide each term in $15 x = 2$ by $15$ and simplify.

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Step 6.2.1

Divide each term in $15 x = 2$ by $15$ .

$\frac{15 x}{15} = \frac{2}{15}$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Cancel the common factor of $15$ .

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Step 6.2.2.1.1

Cancel the common factor.

$\frac{15 x}{15} = \frac{2}{15}$

Step 6.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{15}$

Step 7

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

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Step 7.1

Set the denominator in $\frac{2}{5 x}$ equal to $0$ to find where the expression is undefined.

$5 x = 0$

Step 7.2

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

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Step 7.2.1

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

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

Step 7.2.2

Simplify the left side.

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Step 7.2.2.1

Cancel the common factor of $5$ .

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Step 7.2.2.1.1

Cancel the common factor.

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

Step 7.2.2.1.2

Divide $x$ by $1$ .

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

Step 7.2.3

Simplify the right side.

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Step 7.2.3.1

Divide $0$ by $5$ .

$x = 0$

Step 7.3

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

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

Step 8

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{2}{15}$

$x > \frac{2}{15}$

Step 9

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 9.1

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

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Step 9.1.1

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

$x = - 2$

Step 9.1.2

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

$5 \leq \frac{- 2}{- 5 \cdot - 2} + \frac{4}{2}$

Step 9.1.3

The left side $5$ is greater than the right side $1.8$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval $0 < x < \frac{2}{15}$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $0 < x < \frac{2}{15}$ and see if this value makes the original inequality true.

$x = 0.07$

Step 9.2.2

Replace $x$ with $0.07$ in the original inequality.

$5 \leq \frac{- 2}{- 5 \cdot 0.07} + \frac{4}{2}$

Step 9.2.3

The left side $5$ is less than the right side $7.\overline{714285}$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $x > \frac{2}{15}$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $x > \frac{2}{15}$ and see if this value makes the original inequality true.

$x = 3$

Step 9.3.2

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

$5 \leq \frac{- 2}{- 5 \cdot 3} + \frac{4}{2}$

Step 9.3.3

The left side $5$ is greater than the right side $2.1\overline{3}$ , which means that the given statement is false.

False

Step 9.4

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

$x < 0$ False

$0 < x < \frac{2}{15}$ True

$x > \frac{2}{15}$ False

$x < 0$ False

$0 < x < \frac{2}{15}$ True

$x > \frac{2}{15}$ False

Step 10

The solution consists of all of the true intervals.

$0 < x \leq \frac{2}{15}$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$0 < x \leq \frac{2}{15}$

Interval Notation:

$(0, \frac{2}{15}]$

Step 12