Solve the Inequality for x (5-x)/(2x+1)>=3

$\frac{5 - x}{2 x + 1} \geq 3$

Step 1

Subtract $3$ from both sides of the inequality.

$\frac{5 - x}{2 x + 1} - 3 \geq 0$

Step 2

Simplify $\frac{5 - x}{2 x + 1} - 3$ .

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

$\frac{5 - x}{2 x + 1} - 3 \cdot \frac{2 x + 1}{2 x + 1} \geq 0$

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $2$ by $- 3$ .

$\frac{5 - x - 6 x - 3 \cdot 1}{2 x + 1} \geq 0$

Multiply $- 3$ by $1$ .

$\frac{5 - x - 6 x - 3}{2 x + 1} \geq 0$

Subtract $3$ from $5$ .

$\frac{- x - 6 x + 2}{2 x + 1} \geq 0$

Subtract $6 x$ from $- x$ .

$\frac{- 7 x + 2}{2 x + 1} \geq 0$

Factor $- 1$ out of $- 7 x$ .

$\frac{- (7 x) + 2}{2 x + 1} \geq 0$

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

$\frac{- (7 x) - 1 (- 2)}{2 x + 1} \geq 0$

Factor $- 1$ out of $- (7 x) - 1 (- 2)$ .

$\frac{- (7 x - 2)}{2 x + 1} \geq 0$

Rewrite $- (7 x - 2)$ as $- 1 (7 x - 2)$ .

$\frac{- 1 (7 x - 2)}{2 x + 1} \geq 0$

Move the negative in front of the fraction.

$- \frac{7 x - 2}{2 x + 1} \geq 0$

Step 3

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

$7 x - 2 = 0$

$2 x + 1 = 0$

Step 4

Add $2$ to both sides of the equation.

$7 x = 2$

Step 5

Divide each term in $7 x = 2$ by $7$ and simplify.

Divide each term in $7 x = 2$ by $7$ .

$\frac{7 x}{7} = \frac{2}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{2}{7}$

Divide $x$ by $1$ .

$x = \frac{2}{7}$

Step 6

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 7

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 8

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

$x = \frac{2}{7}$

$x = - \frac{1}{2}$

Step 9

Consolidate the solutions.

$x = \frac{2}{7}, - \frac{1}{2}$

Step 10

Find the domain of $- \frac{7 x - 2}{2 x + 1}$ .

Set the denominator in $\frac{7 x - 2}{2 x + 1}$ equal to $0$ to find where the expression is undefined.

$2 x + 1 = 0$

Solve for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

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

$(- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \infty)$

Step 11

Use each root to create test intervals.

$x < - \frac{1}{2}$

$- \frac{1}{2} < x < \frac{2}{7}$

$x > \frac{2}{7}$

Step 12

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 $x < - \frac{1}{2}$ to see if it makes the inequality true.

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

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

$\frac{5 - (- 3)}{2 (- 3) + 1} \geq 3$

The left side $- 1.6$ is less than the right side $3$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{1}{2} < x < \frac{2}{7}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{1}{2} < x < \frac{2}{7}$ and see if this value makes the original inequality true.

$x = 0$

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

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

The left side $5$ is greater than the right side $3$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{2}{7}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{2}{7}$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

$\frac{5 - (3)}{2 (3) + 1} \geq 3$

The left side $0.\overline{285714}$ is less than the right side $3$ , which means that the given statement is false.

False

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

$x < - \frac{1}{2}$ False

$- \frac{1}{2} < x < \frac{2}{7}$ True

$x > \frac{2}{7}$ False

$x < - \frac{1}{2}$ False

$- \frac{1}{2} < x < \frac{2}{7}$ True

$x > \frac{2}{7}$ False

Step 13

The solution consists of all of the true intervals.

$- \frac{1}{2} < x \leq \frac{2}{7}$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$- \frac{1}{2} < x \leq \frac{2}{7}$

Interval Notation:

$(- \frac{1}{2}, \frac{2}{7}]$

Step 15

Solve the Inequality for x (5-x)/(2x+1)>=3

Related Questions

Support LCMGCF.com by disabling your adblocker.