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

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

Write $| \frac{x + 1}{x - 2} | < 1$ 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.

$x + 1 = 0$

$x - 2 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Add $2$ to 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.

$x - 2 = 0$

Add $2$ to 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.

$x < - 1$

$- 1 < x < 2$

$x > 2$

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 $x < - 1$ 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

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

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

$x = 0$

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

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

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 $x > 2$ 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

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

$x < - 1$ True

$- 1 < x < 2$ False

$x > 2$ True

$x < - 1$ True

$- 1 < x < 2$ False

$x > 2$ True

The solution consists of all of the true intervals.

$x \leq - 1$ or $x > 2$

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

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

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

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.

$x - 2 = 0$

Add $2$ to 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 $x \leq - 1 or x > 2$ and $(- \infty, 2) \cup (2, \infty)$ .

$x \leq - 1 or x > 2$

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.

$x + 1 = 0$

$x - 2 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Add $2$ to 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.

$x - 2 = 0$

Add $2$ to 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.

$x < - 1$

$- 1 < x < 2$

$x > 2$

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 $x < - 1$ 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

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

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

$x = 0$

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

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

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 $x > 2$ 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

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

$x < - 1$ False

$- 1 < x < 2$ True

$x > 2$ False

$x < - 1$ False

$- 1 < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$- 1 < x < 2$

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

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

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

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.

$x - 2 = 0$

Add $2$ to 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 $- 1 < x < 2$ and $(- \infty, 2) \cup (2, \infty)$ .

$- 1 < x < 2$

Write as a piecewise.

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

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

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

Subtract $1$ from both sides of the inequality.

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 1$ by $- 2$ .

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

Subtract $x$ from $x$ .

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

Add $0$ and $1$ .

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

Add $1$ and $2$ .

$\frac{3}{x - 2} < 0$

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

$x - 2 = 0$

Add $2$ to 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.

$x - 2 = 0$

Add $2$ to 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 $x \leq - 1 or x > 2$ .

$x \leq - 1$

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

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

Subtract $1$ from both sides of the inequality.

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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

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

Add $x$ and $x$ .

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

Subtract $2$ from $1$ .

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

Move the negative in front of the fraction.

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

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

$2 x - 1 = 0$

$x - 2 = 0$

Add $1$ to 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 $1$ .

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

Add $2$ to 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.

$x - 2 = 0$

Add $2$ to 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.

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

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

$x > 2$

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

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

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

True

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

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

$x = 1$

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

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

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

False

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.

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

The left side $- 2.5$ 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.

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

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

$x > 2$ True

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

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

$x > 2$ True

The solution consists of all of the true intervals.

$x < \frac{1}{2}$ or $x > 2$

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

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

Find the union of the solutions.

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

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.

$x - 2 = 0$

Add $2$ to 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.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.