Solve the Inequality for x (x+12)/(x+2)-3>=0

$\frac{x + 12}{x + 2} - 3 \geq 0$

Simplify $\frac{x + 12}{x + 2} - 3$ .

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{x + 2}{x + 2}$ .

$\frac{x + 12}{x + 2} - 3 \cdot \frac{x + 2}{x + 2} \geq 0$

Simplify terms.

Combine $- 3$ and $\frac{x + 2}{x + 2}$ .

$\frac{x + 12}{x + 2} + \frac{- 3 (x + 2)}{x + 2} \geq 0$

Combine the numerators over the common denominator.

$\frac{x + 12 - 3 (x + 2)}{x + 2} \geq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{x + 12 - 3 x - 3 \cdot 2}{x + 2} \geq 0$

Multiply $- 3$ by $2$ .

$\frac{x + 12 - 3 x - 6}{x + 2} \geq 0$

Subtract $3 x$ from $x$ .

$\frac{- 2 x + 12 - 6}{x + 2} \geq 0$

Subtract $6$ from $12$ .

$\frac{- 2 x + 6}{x + 2} \geq 0$

Factor $2$ out of $- 2 x + 6$ .

Factor $2$ out of $- 2 x$ .

$\frac{2 (- x) + 6}{x + 2} \geq 0$

Factor $2$ out of $6$ .

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

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

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

Simplify with factoring out.

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

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

Rewrite $3$ as $- 1 (- 3)$ .

$\frac{2 (- (x) - 1 (- 3))}{x + 2} \geq 0$

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

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

Simplify the expression.

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

$\frac{2 (- 1 (x - 3))}{x + 2} \geq 0$

Move the negative in front of the fraction.

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

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

$x - 3 = 0$

$x + 2 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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.

$x = 3$

$x = - 2$

Consolidate the solutions.

$x = 3, - 2$

Find the domain of $\frac{x + 12}{x + 2} - 3$ .

Set the denominator in $\frac{x + 12}{x + 2}$ 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.

$x < - 2$

$- 2 < x < 3$

$x > 3$

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.

$\frac{(- 4) + 12}{(- 4) + 2} - 3 \geq 0$

The left side $- 7$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 2 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{(0) + 12}{(0) + 2} - 3 \geq 0$

The left side $3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{(6) + 12}{(6) + 2} - 3 \geq 0$

The left side $- 0.75$ is less than the right side $0$ , which means that the given statement is false.

False

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

$x < - 2$ False

$- 2 < x < 3$ True

$x > 3$ False

$x < - 2$ False

$- 2 < x < 3$ True

$x > 3$ False

The solution consists of all of the true intervals.

$- 2 < x \leq 3$

The result can be shown in multiple forms.

Inequality Form:

$- 2 < x \leq 3$

Interval Notation:

$(- 2, 3]$

Solve the Inequality for x (x+12)/(x+2)-3>=0

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