Solve the Inequality for x |(x-1)/(x+2)|<=1

$| \frac{x - 1}{x + 2} | \leq 1$

Step 1

Write $| \frac{x - 1}{x + 2} | \leq 1$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

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

Solve the inequality.

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

$x - 1 = 0$

$x + 2 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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.

$x = 1$

$x = - 2$

Consolidate the solutions.

$x = 1, - 2$

Find the domain of $| \frac{x - 1}{x + 2} | - 1$ .

Set the denominator in $\frac{x - 1}{x + 2}$ 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.

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

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 1$

$x > 1$

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.

$\frac{(- 4) - 1}{(- 4) + 2} \geq 0$

The left side $2.5$ is greater than the right side $0$ , 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.

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

The left side $- 0.5$ is less than the right side $0$ , 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.

$\frac{(4) - 1}{(4) + 2} \geq 0$

The left side $0.5$ is greater than the right side $0$ , 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

The solution consists of all of the true intervals.

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

In the piece where $\frac{x - 1}{x + 2}$ is non-negative, remove the absolute value.

$\frac{x - 1}{x + 2} \leq 1$

Find the domain of $\frac{x - 1}{x + 2} \leq 1$ and find the intersection with $x < - 2 or x \geq 1$ .

Find the domain of $\frac{x - 1}{x + 2} \leq 1$ .

Set the denominator in $\frac{x - 1}{x + 2}$ 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.

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

Find the intersection of $x < - 2 or x \geq 1$ and $(- \infty, - 2) \cup (- 2, \infty)$ .

$x < - 2 or x \geq 1$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$\frac{x - 1}{x + 2} < 0$

Solve the inequality.

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

$x - 1 = 0$

$x + 2 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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.

$x = 1$

$x = - 2$

Consolidate the solutions.

$x = 1, - 2$

Find the domain of $| \frac{x - 1}{x + 2} | - 1$ .

Set the denominator in $\frac{x - 1}{x + 2}$ 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.

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

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 1$

$x > 1$

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.

$\frac{(- 4) - 1}{(- 4) + 2} < 0$

The left side $2.5$ is not less than the right side $0$ , which means that the given statement is false.

False

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.

$\frac{(0) - 1}{(0) + 2} < 0$

The left side $- 0.5$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

$\frac{(4) - 1}{(4) + 2} < 0$

The left side $0.5$ is not 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

$- 2 < x < 1$ True

$x > 1$ False

$x < - 2$ False

$- 2 < x < 1$ True

$x > 1$ False

The solution consists of all of the true intervals.

$- 2 < x < 1$

In the piece where $\frac{x - 1}{x + 2}$ is negative, remove the absolute value and multiply by $- 1$ .

$- \frac{x - 1}{x + 2} \leq 1$

Find the domain of $- \frac{x - 1}{x + 2} \leq 1$ and find the intersection with $- 2 < x < 1$ .

Find the domain of $- \frac{x - 1}{x + 2} \leq 1$ .

Set the denominator in $\frac{x - 1}{x + 2}$ 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.

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

Find the intersection of $- 2 < x < 1$ and $(- \infty, - 2) \cup (- 2, \infty)$ .

$- 2 < x < 1$

Write as a piecewise.

${\begin{cases} \frac{x - 1}{x + 2} \leq 1 & x < - 2 or x \geq 1 \\ - \frac{x - 1}{x + 2} \leq 1 & - 2 < x < 1 \end{cases}$

Step 2

Solve $\frac{x - 1}{x + 2} \leq 1$ when $x < - 2 or x \geq 1$ .

Solve $\frac{x - 1}{x + 2} \leq 1$ for $x$ .

Subtract $1$ from both sides of the inequality.

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

Simplify $\frac{x - 1}{x + 2} - 1$ .

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x + 2}{x + 2}$ .

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

Combine $- 1$ and $\frac{x + 2}{x + 2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 1$ by $2$ .

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

Subtract $x$ from $x$ .

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

Subtract $1$ from $0$ .

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

Subtract $2$ from $- 1$ .

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

Move the negative in front of the fraction.

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

Set the denominator in $\frac{3}{x + 2}$ 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.

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

The solution consists of all of the true intervals.

$x > - 2$

Find the intersection of $x > - 2$ and $x < - 2 or x \geq 1$ .

$x \geq 1$

Step 3

Solve $- \frac{x - 1}{x + 2} \leq 1$ when $- 2 < x < 1$ .

Solve $- \frac{x - 1}{x + 2} \leq 1$ for $x$ .

Subtract $1$ from both sides of the inequality.

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

Simplify $- \frac{x - 1}{x + 2} - 1$ .

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x + 2}{x + 2}$ .

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

Combine $- 1$ and $\frac{x + 2}{x + 2}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Factor $- 1$ out of $- (x - 1) - (x + 2)$ .

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

Add $x$ and $x$ .

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

Add $- 1$ and $2$ .

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

Move the negative in front of the fraction.

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

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

$2 x + 1 = 0$

$x + 2 = 0$

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

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.

$x = - \frac{1}{2}$

$x = - 2$

Consolidate the solutions.

$x = - \frac{1}{2}, - 2$

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

Set the denominator in $\frac{2 x + 1}{x + 2}$ 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.

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

Use each root to create test intervals.

$x < - 2$

$- 2 < x < - \frac{1}{2}$

$x > - \frac{1}{2}$

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.

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

The left side $- 2.5$ is less than the right side $1$ , which means that the given statement is always true.

True

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

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

$x = - 1$

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

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

The left side $2$ is greater than the right side $1$ , which means that the given statement is false.

False

Test a value on the interval $x > - \frac{1}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > - \frac{1}{2}$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $0.5$ is less than the right side $1$ , which means that the given statement is always true.

True

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

$x < - 2$ True

$- 2 < x < - \frac{1}{2}$ False

$x > - \frac{1}{2}$ True

$x < - 2$ True

$- 2 < x < - \frac{1}{2}$ False

$x > - \frac{1}{2}$ True

The solution consists of all of the true intervals.

$x < - 2$ or $x \geq - \frac{1}{2}$

Find the intersection of $x < - 2 or x \geq - \frac{1}{2}$ and $- 2 < x < 1$ .

$- \frac{1}{2} \leq x < 1$

Step 4

Find the union of the solutions.

$x \geq - \frac{1}{2}$

Step 5

Find the domain of $| \frac{x - 1}{x + 2} | - 1$ .

Set the denominator in $\frac{x - 1}{x + 2}$ 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 6

The solution consists of all of the true intervals.

$x \geq - \frac{1}{2}$

Step 7

The result can be shown in multiple forms.

Inequality Form:

$x \geq - \frac{1}{2}$

Interval Notation:

$[- \frac{1}{2}, \infty)$

Step 8

Solve the Inequality for x |(x-1)/(x+2)|<=1

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