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

$\frac{| 2 + 3 x |}{2} \geq 5$

Step 1

Multiply both sides by $2$ .

$\frac{| 2 + 3 x |}{2} \cdot 2 \geq 5 \cdot 2$

Step 2

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{| 2 + 3 x |}{2} \cdot 2 \geq 5 \cdot 2$

Rewrite the expression.

$| 2 + 3 x | \geq 5 \cdot 2$

Simplify the right side.

Multiply $5$ by $2$ .

$| 2 + 3 x | \geq 10$

Step 3

Solve for $x$ .

Write $| 2 + 3 x | \geq 10$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$2 + 3 x \geq 0$

Solve the inequality.

Subtract $2$ from both sides of the inequality.

$3 x \geq - 2$

Divide each term in $3 x \geq - 2$ by $3$ and simplify.

Divide each term in $3 x \geq - 2$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

In the piece where $2 + 3 x$ is non-negative, remove the absolute value.

$2 + 3 x \geq 10$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$2 + 3 x < 0$

Solve the inequality.

Subtract $2$ from both sides of the inequality.

$3 x < - 2$

Divide each term in $3 x < - 2$ by $3$ and simplify.

Divide each term in $3 x < - 2$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

In the piece where $2 + 3 x$ is negative, remove the absolute value and multiply by $- 1$ .

$- (2 + 3 x) \geq 10$

Write as a piecewise.

${\begin{cases} 2 + 3 x \geq 10 & x \geq - \frac{2}{3} \\ - (2 + 3 x) \geq 10 & x < - \frac{2}{3} \end{cases}$

Simplify $- (2 + 3 x) \geq 10$ .

Apply the distributive property.

${\begin{cases} 2 + 3 x \geq 10 & x \geq - \frac{2}{3} \\ - 1 \cdot 2 - (3 x) \geq 10 & x < - \frac{2}{3} \end{cases}$

Multiply $- 1$ by $2$ .

${\begin{cases} 2 + 3 x \geq 10 & x \geq - \frac{2}{3} \\ - 2 - (3 x) \geq 10 & x < - \frac{2}{3} \end{cases}$

Multiply $3$ by $- 1$ .

${\begin{cases} 2 + 3 x \geq 10 & x \geq - \frac{2}{3} \\ - 2 - 3 x \geq 10 & x < - \frac{2}{3} \end{cases}$

Solve $2 + 3 x \geq 10$ for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $2$ from both sides of the inequality.

$3 x \geq 10 - 2$

Subtract $2$ from $10$ .

$3 x \geq 8$

Divide each term in $3 x \geq 8$ by $3$ and simplify.

Divide each term in $3 x \geq 8$ by $3$ .

$\frac{3 x}{3} \geq \frac{8}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{8}{3}$

Divide $x$ by $1$ .

$x \geq \frac{8}{3}$

Solve $- 2 - 3 x \geq 10$ for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Add $2$ to both sides of the inequality.

$- 3 x \geq 10 + 2$

Add $10$ and $2$ .

$- 3 x \geq 12$

Divide each term in $- 3 x \geq 12$ by $- 3$ and simplify.

Divide each term in $- 3 x \geq 12$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 3 x}{- 3} \leq \frac{12}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} \leq \frac{12}{- 3}$

Divide $x$ by $1$ .

$x \leq \frac{12}{- 3}$

Simplify the right side.

Divide $12$ by $- 3$ .

$x \leq - 4$

Find the union of the solutions.

$x \leq - 4$ or $x \geq \frac{8}{3}$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 4 or x \geq \frac{8}{3}$

Interval Notation:

$(- \infty, - 4] \cup [\frac{8}{3}, \infty)$

Step 5

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

Related Questions

Support LCMGCF.com by disabling your adblocker.