Solve the Inequality for x 5/(x+3)<5

$\frac{5}{x + 3} < 5$

Subtract $5$ from both sides of the inequality.

$\frac{5}{x + 3} - 5 < 0$

Simplify $\frac{5}{x + 3} - 5$ .

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$\frac{5}{x + 3} - 5 \cdot \frac{x + 3}{x + 3} < 0$

Combine $- 5$ and $\frac{x + 3}{x + 3}$ .

$\frac{5}{x + 3} + \frac{- 5 (x + 3)}{x + 3} < 0$

Combine the numerators over the common denominator.

$\frac{5 - 5 (x + 3)}{x + 3} < 0$

Simplify the numerator.

Factor $5$ out of $5 - 5 (x + 3)$ .

Factor $5$ out of $5$ .

$\frac{5 (1) - 5 (x + 3)}{x + 3} < 0$

Factor $5$ out of $- 5 (x + 3)$ .

$\frac{5 (1) + 5 (- (x + 3))}{x + 3} < 0$

Factor $5$ out of $5 (1) + 5 (- (x + 3))$ .

$\frac{5 (1 - (x + 3))}{x + 3} < 0$

Apply the distributive property.

$\frac{5 (1 - x - 1 \cdot 3)}{x + 3} < 0$

Multiply $- 1$ by $3$ .

$\frac{5 (1 - x - 3)}{x + 3} < 0$

Subtract $3$ from $1$ .

$\frac{5 (- x - 2)}{x + 3} < 0$

Factor $- 1$ out of $- x$ .

$\frac{5 (- (x) - 2)}{x + 3} < 0$

Rewrite $- 2$ as $- 1 (2)$ .

$\frac{5 (- (x) - 1 (2))}{x + 3} < 0$

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

$\frac{5 (- (x + 2))}{x + 3} < 0$

Rewrite $- (x + 2)$ as $- 1 (x + 2)$ .

$\frac{5 (- 1 (x + 2))}{x + 3} < 0$

Move the negative in front of the fraction.

$- \frac{5 (x + 2)}{x + 3} < 0$

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

$x + 2 = 0$

$x + 3 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = - 2$

$x = - 3$

Consolidate the solutions.

$x = - 2, - 3$

Find the domain of $- \frac{5 (x + 2)}{x + 3}$ .

Set the denominator in $\frac{5 (x + 2)}{x + 3}$ equal to $0$ to find where the expression is undefined.

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 2$

$x > - 2$

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.

$\frac{5}{(- 6) + 3} < 5$

The left side $- 1.\overline{6}$ is less than the right side $5$ , which means that the given statement is always true.

True

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

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

$x = - 2.5$

Replace $x$ with $- 2.5$ in the original inequality.

$\frac{5}{(- 2.5) + 3} < 5$

The left side $10$ is not less than the right side $5$ , which means that the given statement is false.

False

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.

$x = 0$

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

$\frac{5}{(0) + 3} < 5$

The left side $1.\overline{6}$ is less than the right side $5$ , which means that the given statement is always true.

True

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

$x < - 3$ True

$- 3 < x < - 2$ False

$x > - 2$ True

$x < - 3$ True

$- 3 < x < - 2$ False

$x > - 2$ True

The solution consists of all of the true intervals.

$x < - 3$ or $x > - 2$

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or x > - 2$

Interval Notation:

$(- \infty, - 3) \cup (- 2, \infty)$

Solve the Inequality for x 5/(x+3)<5

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