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

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

Step 1

Multiply both sides by $3$ .

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

Step 2

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$| 2 x + 1 | \geq 5 \cdot 3$

Simplify the right side.

Multiply $5$ by $3$ .

$| 2 x + 1 | \geq 15$

Step 3

Solve for $x$ .

Write $| 2 x + 1 | \geq 15$ as a piecewise.

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

$2 x + 1 \geq 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$2 x \geq - 1$

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

Divide each term in $2 x \geq - 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

In the piece where $2 x + 1$ is non-negative, remove the absolute value.

$2 x + 1 \geq 15$

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

$2 x + 1 < 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$2 x < - 1$

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.

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

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

$- (2 x + 1) \geq 15$

Write as a piecewise.

${\begin{cases} 2 x + 1 \geq 15 & x \geq - \frac{1}{2} \\ - (2 x + 1) \geq 15 & x < - \frac{1}{2} \end{cases}$

Simplify $- (2 x + 1) \geq 15$ .

Apply the distributive property.

${\begin{cases} 2 x + 1 \geq 15 & x \geq - \frac{1}{2} \\ - (2 x) - 1 \cdot 1 \geq 15 & x < - \frac{1}{2} \end{cases}$

Multiply $2$ by $- 1$ .

${\begin{cases} 2 x + 1 \geq 15 & x \geq - \frac{1}{2} \\ - 2 x - 1 \cdot 1 \geq 15 & x < - \frac{1}{2} \end{cases}$

Multiply $- 1$ by $1$ .

${\begin{cases} 2 x + 1 \geq 15 & x \geq - \frac{1}{2} \\ - 2 x - 1 \geq 15 & x < - \frac{1}{2} \end{cases}$

Solve $2 x + 1 \geq 15$ for $x$ .

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

Subtract $1$ from both sides of the inequality.

$2 x \geq 15 - 1$

Subtract $1$ from $15$ .

$2 x \geq 14$

Divide each term in $2 x \geq 14$ by $2$ and simplify.

Divide each term in $2 x \geq 14$ by $2$ .

$\frac{2 x}{2} \geq \frac{14}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{14}{2}$

Divide $x$ by $1$ .

$x \geq \frac{14}{2}$

Simplify the right side.

Divide $14$ by $2$ .

$x \geq 7$

Solve $- 2 x - 1 \geq 15$ for $x$ .

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

Add $1$ to both sides of the inequality.

$- 2 x \geq 15 + 1$

Add $15$ and $1$ .

$- 2 x \geq 16$

Divide each term in $- 2 x \geq 16$ by $- 2$ and simplify.

Divide each term in $- 2 x \geq 16$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} \leq \frac{16}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} \leq \frac{16}{- 2}$

Divide $x$ by $1$ .

$x \leq \frac{16}{- 2}$

Simplify the right side.

Divide $16$ by $- 2$ .

$x \leq - 8$

Find the union of the solutions.

$x \leq - 8$ or $x \geq 7$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 8 or x \geq 7$

Interval Notation:

$(- \infty, - 8] \cup [7, \infty)$

Step 5

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

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