Algebra Examples

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

Step 1

Subtract $\frac{2}{x - 1}$ from both sides of the inequality.

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

Step 2

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

Tap for more steps...

Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Write each expression with a common denominator of $(x + 1) (x - 1)$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 2.3.1

Multiply $\frac{1}{x + 1}$ by $\frac{x - 1}{x - 1}$ .

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

Step 2.3.2

Multiply $\frac{2}{x - 1}$ by $\frac{x + 1}{x + 1}$ .

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

Step 2.3.3

Reorder the factors of $(x - 1) (x + 1)$ .

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

Step 2.4

Combine the numerators over the common denominator.

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

Step 2.5

Simplify the numerator.

Tap for more steps...

Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Multiply $- 2$ by $1$ .

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

Step 2.5.3

Subtract $2 x$ from $x$ .

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

Step 2.5.4

Subtract $2$ from $- 1$ .

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

Step 2.6

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

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

Step 2.7

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

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

Step 2.8

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

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

Step 2.9

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

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

Step 2.10

Move the negative in front of the fraction.

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

Step 3

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

$x + 3 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 4

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

Add $1$ to both sides of the equation.

$x = 1$

Step 7

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

$x = - 3$

$x = - 1$

$x = 1$

Step 8

Consolidate the solutions.

$x = - 3, - 1, 1$

Step 9

Find the domain of $- \frac{x + 3}{(x + 1) (x - 1)}$ .

Tap for more steps...

Step 9.1

Set the denominator in $\frac{x + 3}{(x + 1) (x - 1)}$ equal to $0$ to find where the expression is undefined.

$(x + 1) (x - 1) = 0$

Step 9.2

Solve for $x$ .

Tap for more steps...

Step 9.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 1 = 0$

$x - 1 = 0$

Step 9.2.2

Set $x + 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 9.2.2.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 9.2.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 9.2.3

Set $x - 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 9.2.3.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 9.2.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 9.2.4

The final solution is all the values that make $(x + 1) (x - 1) = 0$ true.

$x = - 1, 1$

Step 9.3

The domain is all values of $x$ that make the expression defined.

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

Step 10

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$- 1 < x < 1$

$x > 1$

Step 11

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 11.1

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

Tap for more steps...

Step 11.1.1

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

$x = - 6$

Step 11.1.2

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

$\frac{1}{(- 6) + 1} > \frac{2}{(- 6) - 1}$

Step 11.1.3

The left side $- 0.2$ is greater than the right side $- 0.\overline{285714}$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $- 3 < x < - 1$ to see if it makes the inequality true.

Tap for more steps...

Step 11.2.1

Choose a value on the interval $- 3 < x < - 1$ and see if this value makes the original inequality true.

$x = - 2$

Step 11.2.2

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

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

Step 11.2.3

The left side $- 1$ is not greater than the right side $- 0.\overline{6}$ , which means that the given statement is false.

False

Step 11.3

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

Tap for more steps...

Step 11.3.1

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

$x = 0$

Step 11.3.2

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

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

Step 11.3.3

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

True

Step 11.4

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

Tap for more steps...

Step 11.4.1

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

$x = 4$

Step 11.4.2

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

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

Step 11.4.3

The left side $0.2$ is not greater than the right side $0.\overline{6}$ , which means that the given statement is false.

False

Step 11.5

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

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

Step 12

The solution consists of all of the true intervals.

$x < - 3$ or $- 1 < x < 1$

Step 13

Convert the inequality to interval notation.

$(- \infty, - 3) \cup (- 1, 1)$

Step 14