Solve the Inequality for x 1/(2x+5)-5<0

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

Simplify $\frac{1}{2 x + 5} - 5$ .

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{2 x + 5}{2 x + 5}$ .

$\frac{1}{2 x + 5} - 5 \cdot \frac{2 x + 5}{2 x + 5} < 0$

Simplify terms.

Combine $- 5$ and $\frac{2 x + 5}{2 x + 5}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $2$ by $- 5$ .

$\frac{1 - 10 x - 5 \cdot 5}{2 x + 5} < 0$

Multiply $- 5$ by $5$ .

$\frac{1 - 10 x - 25}{2 x + 5} < 0$

Subtract $25$ from $1$ .

$\frac{- 10 x - 24}{2 x + 5} < 0$

Factor $2$ out of $- 10 x - 24$ .

Factor $2$ out of $- 10 x$ .

$\frac{2 (- 5 x) - 24}{2 x + 5} < 0$

Factor $2$ out of $- 24$ .

$\frac{2 (- 5 x) + 2 (- 12)}{2 x + 5} < 0$

Factor $2$ out of $2 (- 5 x) + 2 (- 12)$ .

$\frac{2 (- 5 x - 12)}{2 x + 5} < 0$

Simplify with factoring out.

Factor $- 1$ out of $- 5 x$ .

$\frac{2 (- (5 x) - 12)}{2 x + 5} < 0$

Rewrite $- 12$ as $- 1 (12)$ .

$\frac{2 (- (5 x) - 1 (12))}{2 x + 5} < 0$

Factor $- 1$ out of $- (5 x) - 1 (12)$ .

$\frac{2 (- (5 x + 12))}{2 x + 5} < 0$

Simplify the expression.

Rewrite $- (5 x + 12)$ as $- 1 (5 x + 12)$ .

$\frac{2 (- 1 (5 x + 12))}{2 x + 5} < 0$

Move the negative in front of the fraction.

$- \frac{2 (5 x + 12)}{2 x + 5} < 0$

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

$5 x + 12 = 0$

$2 x + 5 = 0$

Subtract $12$ from both sides of the equation.

$5 x = - 12$

Divide each term in $5 x = - 12$ by $5$ and simplify.

Divide each term in $5 x = - 12$ by $5$ .

$\frac{5 x}{5} = \frac{- 12}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 12}{5}$

Divide $x$ by $1$ .

$x = \frac{- 12}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{12}{5}$

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

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

$x = - \frac{12}{5}$

$x = - \frac{5}{2}$

Consolidate the solutions.

$x = - \frac{12}{5}, - \frac{5}{2}$

Find the domain of $\frac{1}{2 x + 5} - 5$ .

Set the denominator in $\frac{1}{2 x + 5}$ equal to $0$ to find where the expression is undefined.

$2 x + 5 = 0$

Solve for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

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

$(- \infty, - \frac{5}{2}) \cup (- \frac{5}{2}, \infty)$

Use each root to create test intervals.

$x < - \frac{5}{2}$

$- \frac{5}{2} < x < - \frac{12}{5}$

$x > - \frac{12}{5}$

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 $x < - \frac{5}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{5}{2}$ and see if this value makes the original inequality true.

$x = - 5$

Replace $x$ with $- 5$ in the original inequality.

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

The left side $- 5.2$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{5}{2} < x < - \frac{12}{5}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{5}{2} < x < - \frac{12}{5}$ and see if this value makes the original inequality true.

$x = - 2.45$

Replace $x$ with $- 2.45$ in the original inequality.

$\frac{1}{2 (- 2.45) + 5} - 5 < 0$

The left side $5$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > - \frac{12}{5}$ to see if it makes the inequality true.

Choose a value on the interval $x > - \frac{12}{5}$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $- 4.8$ is less 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.

$x < - \frac{5}{2}$ True

$- \frac{5}{2} < x < - \frac{12}{5}$ False

$x > - \frac{12}{5}$ True

$x < - \frac{5}{2}$ True

$- \frac{5}{2} < x < - \frac{12}{5}$ False

$x > - \frac{12}{5}$ True

The solution consists of all of the true intervals.

$x < - \frac{5}{2}$ or $x > - \frac{12}{5}$

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{5}{2} or x > - \frac{12}{5}$

Interval Notation:

$(- \infty, - \frac{5}{2}) \cup (- \frac{12}{5}, \infty)$

Solve the Inequality for x 1/(2x+5)-5<0

Related Questions

Support LCMGCF.com by disabling your adblocker.