Solve the Inequality for x 1/(x+4)<=-2/(1-x)

$\frac{1}{x + 4} \leq \frac{- 2}{1 - x}$

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

$\frac{1}{x + 4} - \frac{- 2}{1 - x} \leq 0$

Simplify $\frac{1}{x + 4} - \frac{- 2}{1 - x}$ .

Simplify each term.

Move the negative in front of the fraction.

$\frac{1}{x + 4} - - \frac{2}{1 - x} \leq 0$

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

Multiply $- 1$ by $- 1$ .

$\frac{1}{x + 4} + 1 \frac{2}{1 - x} \leq 0$

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

$\frac{1}{x + 4} + \frac{2}{1 - x} \leq 0$

To write $\frac{1}{x + 4}$ as a fraction with a common denominator, multiply by $\frac{1 - x}{1 - x}$ .

$\frac{1}{x + 4} \cdot \frac{1 - x}{1 - x} + \frac{2}{1 - x} \leq 0$

To write $\frac{2}{1 - x}$ as a fraction with a common denominator, multiply by $\frac{x + 4}{x + 4}$ .

$\frac{1}{x + 4} \cdot \frac{1 - x}{1 - x} + \frac{2}{1 - x} \cdot \frac{x + 4}{x + 4} \leq 0$

Write each expression with a common denominator of $(x + 4) (1 - x)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{x + 4}$ by $\frac{1 - x}{1 - x}$ .

$\frac{1 - x}{(x + 4) (1 - x)} + \frac{2}{1 - x} \cdot \frac{x + 4}{x + 4} \leq 0$

Multiply $\frac{2}{1 - x}$ by $\frac{x + 4}{x + 4}$ .

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

Reorder the factors of $(x + 4) (1 - x)$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

$\frac{1 - x + 2 x + 2 \cdot 4}{(1 - x) (x + 4)} \leq 0$

Multiply $2$ by $4$ .

$\frac{1 - x + 2 x + 8}{(1 - x) (x + 4)} \leq 0$

Add $1$ and $8$ .

$\frac{- x + 2 x + 9}{(1 - x) (x + 4)} \leq 0$

Add $- x$ and $2 x$ .

$\frac{x + 9}{(1 - x) (x + 4)} \leq 0$

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

$x + 9 = 0$

$1 - x = 0$

$x + 4 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

Subtract $1$ from both sides of the equation.

$- x = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

Subtract $4$ from both sides of the equation.

$x = - 4$

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

$x = - 9$

$x = 1$

$x = - 4$

Consolidate the solutions.

$x = - 9, 1, - 4$

Find the domain of $\frac{x + 9}{(1 - x) (x + 4)}$ .

Set the denominator in $\frac{x + 9}{(1 - x) (x + 4)}$ equal to $0$ to find where the expression is undefined.

$(1 - x) (x + 4) = 0$

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 $0$ .

$1 - x = 0$

$x + 4 = 0$

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

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

$1 - x = 0$

Solve $1 - x = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $(1 - x) (x + 4) = 0$ true.

$x = 1, - 4$

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

$(- \infty, - 4) \cup (- 4, 1) \cup (1, \infty)$

Use each root to create test intervals.

$x < - 9$

$- 9 < x < - 4$

$- 4 < x < 1$

$x > 1$

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 $x < - 9$ to see if it makes the inequality true.

Choose a value on the interval $x < - 9$ and see if this value makes the original inequality true.

$x = - 12$

Replace $x$ with $- 12$ in the original inequality.

$\frac{1}{(- 12) + 4} \leq \frac{- 2}{1 - (- 12)}$

The left side $- 0.125$ is greater than the right side $- 0.\overline{153846}$ , which means that the given statement is false.

False

Test a value on the interval $- 9 < x < - 4$ to see if it makes the inequality true.

Choose a value on the interval $- 9 < x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

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

$\frac{1}{(- 6) + 4} \leq \frac{- 2}{1 - (- 6)}$

The left side $- 0.5$ is less than the right side $- 0.\overline{285714}$ , which means that the given statement is always true.

True

Test a value on the interval $- 4 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 4 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $0.25$ is greater than the right side $- 2$ , 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.

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

The left side $0.125$ is less than the right side $0.\overline{6}$ , which means that the given statement is always true.

True

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

$x < - 9$ False

$- 9 < x < - 4$ True

$- 4 < x < 1$ False

$x > 1$ True

$x < - 9$ False

$- 9 < x < - 4$ True

$- 4 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

$- 9 \leq x < - 4$ or $x > 1$

The result can be shown in multiple forms.

Inequality Form:

$- 9 \leq x < - 4 or x > 1$

Interval Notation:

$[- 9, - 4) \cup (1, \infty)$

Solve the Inequality for x 1/(x+4)<=-2/(1-x)

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