Solve the Inequality for x 15/(x+2)<=5

$\frac{15}{x + 2} \leq 5$

Step 1

Subtract $5$ from both sides of the inequality.

$\frac{15}{x + 2} - 5 \leq 0$

Step 2

Simplify $\frac{15}{x + 2} - 5$ .

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

$\frac{15}{x + 2} - 5 \cdot \frac{x + 2}{x + 2} \leq 0$

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

$\frac{15}{x + 2} + \frac{- 5 (x + 2)}{x + 2} \leq 0$

Combine the numerators over the common denominator.

$\frac{15 - 5 (x + 2)}{x + 2} \leq 0$

Simplify the numerator.

Factor $5$ out of $15 - 5 (x + 2)$ .

Factor $5$ out of $15$ .

$\frac{5 (3) - 5 (x + 2)}{x + 2} \leq 0$

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

$\frac{5 (3) + 5 (- (x + 2))}{x + 2} \leq 0$

Factor $5$ out of $5 (3) + 5 (- (x + 2))$ .

$\frac{5 (3 - (x + 2))}{x + 2} \leq 0$

Apply the distributive property.

$\frac{5 (3 - x - 1 \cdot 2)}{x + 2} \leq 0$

Multiply $- 1$ by $2$ .

$\frac{5 (3 - x - 2)}{x + 2} \leq 0$

Subtract $2$ from $3$ .

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

Factor $- 1$ out of $- x$ .

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

Rewrite $1$ as $- 1 (- 1)$ .

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

Factor $- 1$ out of $- (x) - 1 (- 1)$ .

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

Rewrite $- (x - 1)$ as $- 1 (x - 1)$ .

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

Move the negative in front of the fraction.

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

Step 3

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

$x - 1 = 0$

$x + 2 = 0$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

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.

$x = 1$

$x = - 2$

Step 7

Consolidate the solutions.

$x = 1, - 2$

Step 8

Find the domain of $- \frac{5 (x - 1)}{x + 2}$ .

Set the denominator in $\frac{5 (x - 1)}{x + 2}$ equal to $0$ to find where the expression is undefined.

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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.

$x < - 2$

$- 2 < x < 1$

$x > 1$

Step 10

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 $x < - 2$ to see if it makes the inequality true.

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

$x = - 4$

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

$\frac{15}{(- 4) + 2} \leq 5$

The left side $- 7.5$ is less than the right side $5$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\frac{15}{(0) + 2} \leq 5$

The left side $7.5$ is greater than the right side $5$ , which means that the given statement is false.

False

Test a value on the interval $x > 1$ to see if it makes the inequality true.

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{15}{(4) + 2} \leq 5$

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

True

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

$x < - 2$ True

$- 2 < x < 1$ False

$x > 1$ True

$x < - 2$ True

$- 2 < x < 1$ False

$x > 1$ True

Step 11

The solution consists of all of the true intervals.

$x < - 2$ or $x \geq 1$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < - 2 or x \geq 1$

Interval Notation:

$(- \infty, - 2) \cup [1, \infty)$

Step 13

Solve the Inequality for x 15/(x+2)<=5

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