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

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

Step 1

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

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

Step 2

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

To write $\frac{2}{x + 3}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

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

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

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

Write each expression with a common denominator of $(x + 3) (x + 1)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2}{x + 3}$ by $\frac{x + 1}{x + 1}$ .

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

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

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

Reorder the factors of $(x + 3) (x + 1)$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $3$ .

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

Subtract $x$ from $2 x$ .

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

Subtract $3$ from $2$ .

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

Step 3

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

$x - 1 = 0$

$x + 1 = 0$

$x + 3 = 0$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 7

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

$x = 1$

$x = - 1$

$x = - 3$

Step 8

Consolidate the solutions.

$x = 1, - 1, - 3$

Step 9

Find the domain of $\frac{x - 1}{(x + 1) (x + 3)}$ .

Set the denominator in $\frac{x - 1}{(x + 1) (x + 3)}$ equal to $0$ to find where the expression is undefined.

$(x + 1) (x + 3) = 0$

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 $0$ .

$x + 1 = 0$

$x + 3 = 0$

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 $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

The final solution is all the values that make $(x + 1) (x + 3) = 0$ true.

$x = - 1, - 3$

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

$(- \infty, - 3) \cup (- 3, - 1) \cup (- 1, \infty)$

Step 10

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$- 1 < x < 1$

$x > 1$

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 $x < - 3$ to see if it makes the inequality true.

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

$x = - 6$

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

$\frac{2}{(- 6) + 3} < \frac{1}{(- 6) + 1}$

The left side $- 0.\overline{6}$ is less than the right side $- 0.2$ , which means that the given statement is always true.

True

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

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

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

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

The left side $2$ is not less than the right side $- 1$ , which means that the given statement is false.

False

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.

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

The left side $0.\overline{6}$ is less than the right side $1$ , 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.

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

The left side $0.\overline{285714}$ is not less than the right side $0.2$ , which means that the given statement is false.

False

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

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

Step 12

The solution consists of all of the true intervals.

$x < - 3$ or $- 1 < x < 1$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or - 1 < x < 1$

Interval Notation:

$(- \infty, - 3) \cup (- 1, 1)$

Step 14

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

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