Solve the Inequality for x 2/(x+1)<=3/(x-1)

$\frac{2}{x + 1} \leq \frac{3}{x - 1}$

Subtract $\frac{3}{x - 1}$ from both sides of the inequality.

$\frac{2}{x + 1} - \frac{3}{x - 1} \leq 0$

Simplify $\frac{2}{x + 1} - \frac{3}{x - 1}$ .

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

$\frac{2}{x + 1} \cdot \frac{x - 1}{x - 1} - \frac{3}{x - 1} \leq 0$

To write $- \frac{3}{x - 1}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

$\frac{2}{x + 1} \cdot \frac{x - 1}{x - 1} - \frac{3}{x - 1} \cdot \frac{x + 1}{x + 1} \leq 0$

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

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

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

Multiply $\frac{3}{x - 1}$ by $\frac{x + 1}{x + 1}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

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

Apply the distributive property.

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

Multiply $- 3$ by $1$ .

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

Subtract $3 x$ from $2 x$ .

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

Subtract $3$ from $- 2$ .

$\frac{- x - 5}{(x + 1) (x - 1)} \leq 0$

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

$\frac{- (x) - 5}{(x + 1) (x - 1)} \leq 0$

Rewrite $- 5$ as $- 1 (5)$ .

$\frac{- (x) - 1 (5)}{(x + 1) (x - 1)} \leq 0$

Factor $- 1$ out of $- (x) - 1 (5)$ .

$\frac{- (x + 5)}{(x + 1) (x - 1)} \leq 0$

Rewrite $- (x + 5)$ as $- 1 (x + 5)$ .

$\frac{- 1 (x + 5)}{(x + 1) (x - 1)} \leq 0$

Move the negative in front of the fraction.

$- \frac{x + 5}{(x + 1) (x - 1)} \leq 0$

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

$x + 5 = 0$

$x + 1 = 0$

$x - 1 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Subtract $1$ from both sides of the equation.

$x = - 1$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = - 5$

$x = - 1$

$x = 1$

Consolidate the solutions.

$x = - 5, - 1, 1$

Find the domain of $- \frac{x + 5}{(x + 1) (x - 1)}$ .

Set the denominator in $\frac{x + 5}{(x + 1) (x - 1)}$ equal to $0$ to find where the expression is undefined.

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

Solve for $x$ .

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

$x + 1 = 0$

$x - 1 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = - 1, 1$

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

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

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 1$

$- 1 < x < 1$

$x > 1$

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 $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$\frac{2}{(- 8) + 1} \leq \frac{3}{(- 8) - 1}$

The left side $- 0.\overline{285714}$ is greater than the right side $- 0.\overline{3}$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < - 1$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

$\frac{2}{(- 3) + 1} \leq \frac{3}{(- 3) - 1}$

The left side $- 1$ is less than the right side $- 0.75$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

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

The left side $2$ is greater than the right side $- 3$ , 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.

$\frac{2}{(4) + 1} \leq \frac{3}{(4) - 1}$

The left side $0.4$ 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.

$x < - 5$ False

$- 5 < x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

$x < - 5$ False

$- 5 < x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$- 5 \leq x < - 1$ or $x > 1$

The result can be shown in multiple forms.

Inequality Form:

$- 5 \leq x < - 1 or x > 1$

Interval Notation:

$[- 5, - 1) \cup (1, \infty)$

Solve the Inequality for x 2/(x+1)<=3/(x-1)

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