Solve for x |2/3x-2|>8

$| \frac{2}{3} x - 2 | > 8$

Step 1

Write $| \frac{2}{3} x - 2 | > 8$ as a piecewise.

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

$\frac{2}{3} x - 2 \geq 0$

Solve the inequality.

Combine $\frac{2}{3}$ and $x$ .

$\frac{2 x}{3} - 2 \geq 0$

Add $2$ to both sides of the inequality.

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

Multiply both sides by $3$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$2 x \geq 2 \cdot 3$

Simplify the right side.

Multiply $2$ by $3$ .

$2 x \geq 6$

Divide each term in $2 x \geq 6$ by $2$ and simplify.

Divide each term in $2 x \geq 6$ by $2$ .

$\frac{2 x}{2} \geq \frac{6}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \geq \frac{6}{2}$

Divide $x$ by $1$ .

$x \geq \frac{6}{2}$

Simplify the right side.

Divide $6$ by $2$ .

$x \geq 3$

In the piece where $\frac{2}{3} x - 2$ is non-negative, remove the absolute value.

$\frac{2}{3} x - 2 > 8$

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

$\frac{2}{3} x - 2 < 0$

Solve the inequality.

Combine $\frac{2}{3}$ and $x$ .

$\frac{2 x}{3} - 2 < 0$

Add $2$ to both sides of the inequality.

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

Multiply both sides by $3$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$2 x < 2 \cdot 3$

Simplify the right side.

Multiply $2$ by $3$ .

$2 x < 6$

Divide each term in $2 x < 6$ by $2$ and simplify.

Divide each term in $2 x < 6$ by $2$ .

$\frac{2 x}{2} < \frac{6}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} < \frac{6}{2}$

Divide $x$ by $1$ .

$x < \frac{6}{2}$

Simplify the right side.

Divide $6$ by $2$ .

$x < 3$

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

$- (\frac{2}{3} x - 2) > 8$

Write as a piecewise.

${\begin{cases} \frac{2}{3} x - 2 > 8 & x \geq 3 \\ - (\frac{2}{3} x - 2) > 8 & x < 3 \end{cases}$

Combine $\frac{2}{3}$ and $x$ .

${\begin{cases} \frac{2 x}{3} - 2 > 8 & x \geq 3 \\ - (\frac{2}{3} x - 2) > 8 & x < 3 \end{cases}$

Simplify $- (\frac{2}{3} x - 2) > 8$ .

Combine $\frac{2}{3}$ and $x$ .

${\begin{cases} \frac{2 x}{3} - 2 > 8 & x \geq 3 \\ - (\frac{2 x}{3} - 2) > 8 & x < 3 \end{cases}$

Apply the distributive property.

${\begin{cases} \frac{2 x}{3} - 2 > 8 & x \geq 3 \\ - \frac{2 x}{3} - - 2 > 8 & x < 3 \end{cases}$

Multiply $- 1$ by $- 2$ .

${\begin{cases} \frac{2 x}{3} - 2 > 8 & x \geq 3 \\ - \frac{2 x}{3} + 2 > 8 & x < 3 \end{cases}$

Step 2

Solve $\frac{2 x}{3} - 2 > 8$ for $x$ .

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

Add $2$ to both sides of the inequality.

$\frac{2 x}{3} > 8 + 2$

Add $8$ and $2$ .

$\frac{2 x}{3} > 10$

Multiply both sides by $3$ .

$\frac{2 x}{3} \cdot 3 > 10 \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{2 x}{3} \cdot 3 > 10 \cdot 3$

Rewrite the expression.

$2 x > 10 \cdot 3$

Simplify the right side.

Multiply $10$ by $3$ .

$2 x > 30$

Divide each term in $2 x > 30$ by $2$ and simplify.

Divide each term in $2 x > 30$ by $2$ .

$\frac{2 x}{2} > \frac{30}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} > \frac{30}{2}$

Divide $x$ by $1$ .

$x > \frac{30}{2}$

Simplify the right side.

Divide $30$ by $2$ .

$x > 15$

Step 3

Solve $- \frac{2 x}{3} + 2 > 8$ for $x$ .

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

Subtract $2$ from both sides of the inequality.

$- \frac{2 x}{3} > 8 - 2$

Subtract $2$ from $8$ .

$- \frac{2 x}{3} > 6$

Divide each term in $- \frac{2 x}{3} > 6$ by $- 1$ and simplify.

Divide each term in $- \frac{2 x}{3} > 6$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \frac{2 x}{3}}{- 1} < \frac{6}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\frac{2 x}{3}}{1} < \frac{6}{- 1}$

Divide $\frac{2 x}{3}$ by $1$ .

$\frac{2 x}{3} < \frac{6}{- 1}$

Simplify the right side.

Divide $6$ by $- 1$ .

$\frac{2 x}{3} < - 6$

Multiply both sides by $3$ .

$\frac{2 x}{3} \cdot 3 < - 6 \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{2 x}{3} \cdot 3 < - 6 \cdot 3$

Rewrite the expression.

$2 x < - 6 \cdot 3$

Simplify the right side.

Multiply $- 6$ by $3$ .

$2 x < - 18$

Divide each term in $2 x < - 18$ by $2$ and simplify.

Divide each term in $2 x < - 18$ by $2$ .

$\frac{2 x}{2} < \frac{- 18}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} < \frac{- 18}{2}$

Divide $x$ by $1$ .

$x < \frac{- 18}{2}$

Simplify the right side.

Divide $- 18$ by $2$ .

$x < - 9$

Step 4

Find the union of the solutions.

$x < - 9$ or $x > 15$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x < - 9 or x > 15$

Interval Notation:

$(- \infty, - 9) \cup (15, \infty)$

Step 6

Solve for x |2/3x-2|>8

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