Solve the Inequality for x 4/(x+2)>2

$\frac{4}{x + 2} > 2$

Subtract $2$ from both sides of the inequality.

$\frac{4}{x + 2} - 2 > 0$

Simplify $\frac{4}{x + 2} - 2$ .

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

$\frac{4}{x + 2} - 2 \cdot \frac{x + 2}{x + 2} > 0$

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

$\frac{4}{x + 2} + \frac{- 2 (x + 2)}{x + 2} > 0$

Combine the numerators over the common denominator.

$\frac{4 - 2 (x + 2)}{x + 2} > 0$

Simplify the numerator.

Factor $2$ out of $4 - 2 (x + 2)$ .

Factor $2$ out of $4$ .

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

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

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

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

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

Apply the distributive property.

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

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $2$ .

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

Add $- x$ and $0$ .

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

Combine exponents.

Factor out negative.

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

Multiply $2$ by $- 1$ .

$\frac{- 2 x}{x + 2} > 0$

Move the negative in front of the fraction.

$- \frac{2 x}{x + 2} > 0$

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

$- 2 x = 0$

$x + 2 = 0$

Divide each term in $- 2 x = 0$ by $- 2$ and simplify.

Divide each term in $- 2 x = 0$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{0}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{0}{- 2}$

Divide $x$ by $1$ .

$x = \frac{0}{- 2}$

Simplify the right side.

Divide $0$ by $- 2$ .

$x = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$x = 0$

$x = - 2$

Consolidate the solutions.

$x = 0, - 2$

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

Set the denominator in $\frac{2 x}{x + 2}$ 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.

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

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 0$

$x > 0$

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.

$\frac{4}{(- 4) + 2} > 2$

The left side $- 2$ is not greater than the right side $2$ , which means that the given statement is false.

False

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

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

$x = - 1$

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

$\frac{4}{(- 1) + 2} > 2$

The left side $4$ is greater than the right side $2$ , which means that the given statement is always true.

True

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

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

$x = 2$

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

$\frac{4}{(2) + 2} > 2$

The left side $1$ is not greater than the right side $2$ , which means that the given statement is false.

False

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

$x < - 2$ False

$- 2 < x < 0$ True

$x > 0$ False

$x < - 2$ False

$- 2 < x < 0$ True

$x > 0$ False

The solution consists of all of the true intervals.

$- 2 < x < 0$

The result can be shown in multiple forms.

Inequality Form:

$- 2 < x < 0$

Interval Notation:

$(- 2, 0)$

Solve the Inequality for x 4/(x+2)>2

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