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

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

Step 1

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

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

Step 2

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Add $1$ and $2$ .

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $x$ by $x$ .

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

Apply the distributive property.

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

Multiply $- 2$ by $1$ .

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

Subtract $2 x$ from $3 x$ .

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

Factor $x^{2} + x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $1$ .

$- 1, 2$

Write the factored form using these integers.

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

Step 3

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

$x = 0$

$x - 1 = 0$

$x + 2 = 0$

$x + 1 = 0$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6

Subtract $1$ from 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.

$x = 0$

$x = 1$

$x = - 2$

$x = - 1$

Step 8

Consolidate the solutions.

$x = 0, 1, - 2, - 1$

Step 9

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

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

$x (x + 1) = 0$

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 $0$ .

$x = 0$

$x + 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

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.

$x = - 1$

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

$x = 0, - 1$

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

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Step 10

Use each root to create test intervals.

$x < - 2$

$- 2 < x < - 1$

$- 1 < x < 0$

$0 < x < 1$

$x > 1$

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.

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

The left side $0.\overline{3}$ is greater than the right side $- 0.5$ , which means that the given statement is false.

False

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

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

$x = - 1.5$

Replace $x$ with $- 1.5$ in the original inequality.

$1 + \frac{2}{(- 1.5) + 1} \leq \frac{2}{- 1.5}$

The left side $- 3$ is less than the right side $- 1.\overline{3}$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Replace $x$ with $- 0.5$ in the original inequality.

$1 + \frac{2}{(- 0.5) + 1} \leq \frac{2}{- 0.5}$

The left side $5$ is greater than the right side $- 4$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Replace $x$ with $0.5$ in the original inequality.

$1 + \frac{2}{(0.5) + 1} \leq \frac{2}{0.5}$

The left side $2.\overline{3}$ is less than the right side $4$ , 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.

$x = 4$

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

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

The left side $1.4$ is greater than the right side $0.5$ , which means that the given statement is false.

False

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

$x < - 2$ False

$- 2 < x < - 1$ True

$- 1 < x < 0$ False

$0 < x < 1$ True

$x > 1$ False

$x < - 2$ False

$- 2 < x < - 1$ True

$- 1 < x < 0$ False

$0 < x < 1$ True

$x > 1$ False

Step 12

The solution consists of all of the true intervals.

$- 2 \leq x < - 1$ or $0 < x \leq 1$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$- 2 \leq x < - 1 or 0 < x \leq 1$

Interval Notation:

$[- 2, - 1) \cup (0, 1]$

Step 14

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

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