Solve the Inequality for x 1/(x+1)+1/(x+2)<=0

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

Step 1

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

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

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

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

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

Write each expression with a common denominator of $(x + 1) (x + 2)$ , by multiplying each by an appropriate factor of $1$ .

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

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

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

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

Reorder the factors of $(x + 2) (x + 1)$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Add $x$ and $x$ .

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

Add $2$ and $1$ .

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

Step 2

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

$2 x + 3 = 0$

$x + 1 = 0$

$x + 2 = 0$

Step 3

Subtract $3$ from both sides of the equation.

$2 x = - 3$

Step 4

Divide each term in $2 x = - 3$ by $2$ and simplify.

Divide each term in $2 x = - 3$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2}$

Divide $x$ by $1$ .

$x = \frac{- 3}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{2}$

Step 5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7

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

$x = - \frac{3}{2}$

$x = - 1$

$x = - 2$

Step 8

Consolidate the solutions.

$x = - \frac{3}{2}, - 1, - 2$

Step 9

Find the domain of $\frac{1}{x + 1} + \frac{1}{x + 2}$ .

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set the denominator in $\frac{1}{x + 2}$ equal to $0$ to find where the expression is undefined.

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$(- \infty, - 2) \cup (- 2, - 1) \cup (- 1, \infty)$

Step 10

Use each root to create test intervals.

$x < - 2$

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

$- \frac{3}{2} < x < - 1$

$x > - 1$

Step 11

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

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.

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

The left side $- 0.8\overline{3}$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 2 < x < - \frac{3}{2}$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < - \frac{3}{2}$ and see if this value makes the original inequality true.

$x = - 1.75$

Replace $x$ with $- 1.75$ in the original inequality.

$\frac{1}{(- 1.75) + 1} + \frac{1}{(- 1.75) + 2} \leq 0$

The left side $2.\overline{6}$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{3}{2} < x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{3}{2} < x < - 1$ and see if this value makes the original inequality true.

$x = - 1.25$

Replace $x$ with $- 1.25$ in the original inequality.

$\frac{1}{(- 1.25) + 1} + \frac{1}{(- 1.25) + 2} \leq 0$

The left side $- 2.\overline{6}$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

$x = 0$

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

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

The left side $1.5$ is greater 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.

$x < - 2$ True

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

$- \frac{3}{2} < x < - 1$ True

$x > - 1$ False

$x < - 2$ True

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

$- \frac{3}{2} < x < - 1$ True

$x > - 1$ False

Step 12

The solution consists of all of the true intervals.

$x < - 2$ or $- \frac{3}{2} \leq x < - 1$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x < - 2 or - \frac{3}{2} \leq x < - 1$

Interval Notation:

$(- \infty, - 2) \cup [- \frac{3}{2}, - 1)$

Step 14

Solve the Inequality for x 1/(x+1)+1/(x+2)<=0

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