Solve the Inequality for x (x+3)/x+1>0

$\frac{x + 3}{x} + 1 > 0$

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

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

$\frac{x + 3}{x} + \frac{x}{x} > 0$

Combine the numerators over the common denominator.

$\frac{x + 3 + x}{x} > 0$

Add $x$ and $x$ .

$\frac{2 x + 3}{x} > 0$

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

$x = 0$

$2 x + 3 = 0$

Subtract $3$ from both sides of the equation.

$2 x = - 3$

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.

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

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

$x = 0$

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

Consolidate the solutions.

$x = 0, - \frac{3}{2}$

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

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

$x = 0$

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

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

Use each root to create test intervals.

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

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

$x > 0$

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 $x < - \frac{3}{2}$ to see if it makes the inequality true.

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

$x = - 4$

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

$\frac{(- 4) + 3}{- 4} + 1 > 0$

The left side $1.25$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{3}{2} < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{3}{2} < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

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

The left side $- 1$ is not greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 2$

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

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

The left side $3.5$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$- \frac{3}{2} < x < 0$ False

$x > 0$ True

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

$- \frac{3}{2} < x < 0$ False

$x > 0$ True

The solution consists of all of the true intervals.

$x < - \frac{3}{2}$ or $x > 0$

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{3}{2} or x > 0$

Interval Notation:

$(- \infty, - \frac{3}{2}) \cup (0, \infty)$

Solve the Inequality for x (x+3)/x+1>0

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