Solve the Inequality for x -2/(7-x)>=1/(x+1)

$\frac{- 2}{7 - x} \geq \frac{1}{x + 1}$

Step 1

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

$\frac{- 2}{7 - x} - \frac{1}{x + 1} \geq 0$

Step 2

Simplify $\frac{- 2}{7 - x} - \frac{1}{x + 1}$ .

Move the negative in front of the fraction.

$- \frac{2}{7 - x} - \frac{1}{x + 1} \geq 0$

To write $- \frac{2}{7 - x}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

$- \frac{2}{7 - x} \cdot \frac{x + 1}{x + 1} - \frac{1}{x + 1} \geq 0$

To write $- \frac{1}{x + 1}$ as a fraction with a common denominator, multiply by $\frac{7 - x}{7 - x}$ .

$- \frac{2}{7 - x} \cdot \frac{x + 1}{x + 1} - \frac{1}{x + 1} \cdot \frac{7 - x}{7 - x} \geq 0$

Write each expression with a common denominator of $(7 - x) (x + 1)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2}{7 - x}$ by $\frac{x + 1}{x + 1}$ .

$- \frac{2 (x + 1)}{(7 - x) (x + 1)} - \frac{1}{x + 1} \cdot \frac{7 - x}{7 - x} \geq 0$

Multiply $\frac{1}{x + 1}$ by $\frac{7 - x}{7 - x}$ .

$- \frac{2 (x + 1)}{(7 - x) (x + 1)} - \frac{7 - x}{(x + 1) (7 - x)} \geq 0$

Reorder the factors of $(x + 1) (7 - x)$ .

$- \frac{2 (x + 1)}{(7 - x) (x + 1)} - \frac{7 - x}{(7 - x) (x + 1)} \geq 0$

Combine the numerators over the common denominator.

$\frac{- 2 (x + 1) - (7 - x)}{(7 - x) (x + 1)} \geq 0$

Simplify the numerator.

Factor $- 1$ out of $- 2 (x + 1) - (7 - x)$ .

Reorder $7$ and $- x$ .

$\frac{- 2 (x + 1) - (- x + 7)}{(7 - x) (x + 1)} \geq 0$

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

$\frac{- (2 (x + 1)) - (- x + 7)}{(7 - x) (x + 1)} \geq 0$

Factor $- 1$ out of $- (2 (x + 1)) - (- x + 7)$ .

$\frac{- (2 (x + 1) - x + 7)}{(7 - x) (x + 1)} \geq 0$

Apply the distributive property.

$\frac{- (2 x + 2 \cdot 1 - x + 7)}{(7 - x) (x + 1)} \geq 0$

Multiply $2$ by $1$ .

$\frac{- (2 x + 2 - x + 7)}{(7 - x) (x + 1)} \geq 0$

Subtract $x$ from $2 x$ .

$\frac{- (x + 2 + 7)}{(7 - x) (x + 1)} \geq 0$

Add $2$ and $7$ .

$\frac{- (x + 9)}{(7 - x) (x + 1)} \geq 0$

Move the negative in front of the fraction.

$- \frac{x + 9}{(7 - x) (x + 1)} \geq 0$

Step 3

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

$x + 9 = 0$

$7 - x = 0$

$x + 1 = 0$

Step 4

Subtract $9$ from both sides of the equation.

$x = - 9$

Step 5

Subtract $7$ from both sides of the equation.

$- x = - 7$

Step 6

Divide each term in $- x = - 7$ by $- 1$ and simplify.

Divide each term in $- x = - 7$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 7}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 7}{- 1}$

Simplify the right side.

Divide $- 7$ by $- 1$ .

$x = 7$

Step 7

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8

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

$x = - 9$

$x = 7$

$x = - 1$

Step 9

Consolidate the solutions.

$x = - 9, 7, - 1$

Step 10

Find the domain of $- \frac{x + 9}{(7 - x) (x + 1)}$ .

Set the denominator in $\frac{x + 9}{(7 - x) (x + 1)}$ equal to $0$ to find where the expression is undefined.

$(7 - x) (x + 1) = 0$

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 $0$ .

$7 - x = 0$

$x + 1 = 0$

Set $7 - x$ equal to $0$ and solve for $x$ .

Set $7 - x$ equal to $0$ .

$7 - x = 0$

Solve $7 - x = 0$ for $x$ .

Subtract $7$ from both sides of the equation.

$- x = - 7$

Divide each term in $- x = - 7$ by $- 1$ and simplify.

Divide each term in $- x = - 7$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 7}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 7}{- 1}$

Simplify the right side.

Divide $- 7$ by $- 1$ .

$x = 7$

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.

$x = - 1$

The final solution is all the values that make $(7 - x) (x + 1) = 0$ true.

$x = 7, - 1$

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

$(- \infty, - 1) \cup (- 1, 7) \cup (7, \infty)$

Step 11

Use each root to create test intervals.

$x < - 9$

$- 9 < x < - 1$

$- 1 < x < 7$

$x > 7$

Step 12

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.

$\frac{- 2}{7 - (- 12)} \geq \frac{1}{(- 12) + 1}$

The left side $- 0.10526315$ is less than the right side $- 0.\overline{09}$ , which means that the given statement is false.

False

Test a value on the interval $- 9 < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- 9 < x < - 1$ and see if this value makes the original inequality true.

$x = - 5$

Replace $x$ with $- 5$ in the original inequality.

$\frac{- 2}{7 - (- 5)} \geq \frac{1}{(- 5) + 1}$

The left side $- 0.1\overline{6}$ is greater than the right side $- 0.25$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 7$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 7$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{- 2}{7 - (0)} \geq \frac{1}{(0) + 1}$

The left side $- 0.\overline{285714}$ is less than the right side $1$ , which means that the given statement is false.

False

Test a value on the interval $x > 7$ to see if it makes the inequality true.

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$\frac{- 2}{7 - (10)} \geq \frac{1}{(10) + 1}$

The left side $0.\overline{6}$ is greater than the right side $0.\overline{09}$ , which means that the given statement is always true.

True

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

$x < - 9$ False

$- 9 < x < - 1$ True

$- 1 < x < 7$ False

$x > 7$ True

$x < - 9$ False

$- 9 < x < - 1$ True

$- 1 < x < 7$ False

$x > 7$ True

Step 13

The solution consists of all of the true intervals.

$- 9 \leq x < - 1$ or $x > 7$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$- 9 \leq x < - 1 or x > 7$

Interval Notation:

$[- 9, - 1) \cup (7, \infty)$

Step 15

Solve the Inequality for x -2/(7-x)>=1/(x+1)

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