Algebra Examples

$\frac{10}{x + 2} \leq 5$

Step 1

Subtract $5$ from both sides of the inequality.

$\frac{10}{x + 2} - 5 \leq 0$

Step 2

Simplify $\frac{10}{x + 2} - 5$ .

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Step 2.1

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{x + 2}{x + 2}$ .

$\frac{10}{x + 2} - 5 \cdot \frac{x + 2}{x + 2} \leq 0$

Step 2.2

Combine $- 5$ and $\frac{x + 2}{x + 2}$ .

$\frac{10}{x + 2} + \frac{- 5 (x + 2)}{x + 2} \leq 0$

Step 2.3

Combine the numerators over the common denominator.

$\frac{10 - 5 (x + 2)}{x + 2} \leq 0$

Step 2.4

Simplify the numerator.

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Step 2.4.1

Factor $5$ out of $10 - 5 (x + 2)$ .

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Step 2.4.1.1

Factor $5$ out of $10$ .

$\frac{5 (2) - 5 (x + 2)}{x + 2} \leq 0$

Step 2.4.1.2

Factor $5$ out of $- 5 (x + 2)$ .

$\frac{5 (2) + 5 (- (x + 2))}{x + 2} \leq 0$

Step 2.4.1.3

Factor $5$ out of $5 (2) + 5 (- (x + 2))$ .

$\frac{5 (2 - (x + 2))}{x + 2} \leq 0$

Step 2.4.2

Apply the distributive property.

$\frac{5 (2 - x - 1 \cdot 2)}{x + 2} \leq 0$

Step 2.4.3

Multiply $- 1$ by $2$ .

$\frac{5 (2 - x - 2)}{x + 2} \leq 0$

Step 2.4.4

Subtract $2$ from $2$ .

$\frac{5 (- x + 0)}{x + 2} \leq 0$

Step 2.4.5

Add $- x$ and $0$ .

$\frac{5 \cdot - 1 x}{x + 2} \leq 0$

Step 2.4.6

Combine exponents.

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Step 2.4.6.1

Factor out negative.

$\frac{- (5 x)}{x + 2} \leq 0$

Step 2.4.6.2

Multiply $5$ by $- 1$ .

$\frac{- 5 x}{x + 2} \leq 0$

Step 2.5

Move the negative in front of the fraction.

$- \frac{5 x}{x + 2} \leq 0$

Step 3

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

$- 5 x = 0$

$x + 2 = 0$

Step 4

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

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Step 4.1

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

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

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 5$ .

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Step 4.2.1.1

Cancel the common factor.

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

Step 4.2.1.2

Divide $x$ by $1$ .

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

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 5$ .

$x = 0$

Step 5

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6

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

$x = 0$

$x = - 2$

Step 7

Consolidate the solutions.

$x = 0, - 2$

Step 8

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

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Step 8.1

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

$x + 2 = 0$

Step 8.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 8.3

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

$(- \infty, - 2) \cup (- 2, \infty)$

Step 9

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 0$

$x > 0$

Step 10

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 10.1

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 10.1.2

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

$\frac{10}{(- 4) + 2} \leq 5$

Step 10.1.3

The left side $- 5$ is less than the right side $5$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $- 2 < x < 0$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Step 10.2.2

Replace $x$ with $- 1$ in the original inequality.

$\frac{10}{(- 1) + 2} \leq 5$

Step 10.2.3

The left side $10$ is greater than the right side $5$ , which means that the given statement is false.

False

Step 10.3

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

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Step 10.3.1

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

$x = 2$

Step 10.3.2

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

$\frac{10}{(2) + 2} \leq 5$

Step 10.3.3

The left side $2.5$ is less than the right side $5$ , which means that the given statement is always true.

True

Step 10.4

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

$x < - 2$ True

$- 2 < x < 0$ False

$x > 0$ True

$x < - 2$ True

$- 2 < x < 0$ False

$x > 0$ True

Step 11

The solution consists of all of the true intervals.

$x < - 2$ or $x \geq 0$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < - 2 or x \geq 0$

Interval Notation:

$(- \infty, - 2) \cup [0, \infty)$

Step 13