Solve the Absolute Value Inequality for x -8|x/3+1|>-8

$- 8 | \frac{x}{3} + 1 | > - 8$

Step 1

Divide each term in $- 8 | \frac{x}{3} + 1 | > - 8$ by $- 8$ and simplify.

Divide each term in $- 8 | \frac{x}{3} + 1 | > - 8$ by $- 8$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 8 | \frac{x}{3} + 1 |}{- 8} < \frac{- 8}{- 8}$

Simplify the left side.

Cancel the common factor of $- 8$ .

Cancel the common factor.

$\frac{- 8 | \frac{x}{3} + 1 |}{- 8} < \frac{- 8}{- 8}$

Divide $| \frac{x}{3} + 1 |$ by $1$ .

$| \frac{x}{3} + 1 | < \frac{- 8}{- 8}$

Simplify the right side.

Divide $- 8$ by $- 8$ .

$| \frac{x}{3} + 1 | < 1$

Step 2

Write $| \frac{x}{3} + 1 | < 1$ as a piecewise.

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

$\frac{x}{3} + 1 \geq 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

$\frac{x}{3} \geq - 1$

Multiply both sides by $3$ .

$\frac{x}{3} \cdot 3 \geq - 1 \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{x}{3} \cdot 3 \geq - 1 \cdot 3$

Rewrite the expression.

$x \geq - 1 \cdot 3$

Simplify the right side.

Multiply $- 1$ by $3$ .

$x \geq - 3$

In the piece where $\frac{x}{3} + 1$ is non-negative, remove the absolute value.

$\frac{x}{3} + 1 < 1$

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

$\frac{x}{3} + 1 < 0$

Solve the inequality.

Subtract $1$ from both sides of the inequality.

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

Multiply both sides by $3$ .

$\frac{x}{3} \cdot 3 < - 1 \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{x}{3} \cdot 3 < - 1 \cdot 3$

Rewrite the expression.

$x < - 1 \cdot 3$

Simplify the right side.

Multiply $- 1$ by $3$ .

$x < - 3$

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

$- (\frac{x}{3} + 1) < 1$

Write as a piecewise.

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

Simplify $- (\frac{x}{3} + 1) < 1$ .

Apply the distributive property.

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

Multiply $- 1$ by $1$ .

${\begin{cases} \frac{x}{3} + 1 < 1 & x \geq - 3 \\ - \frac{x}{3} - 1 < 1 & x < - 3 \end{cases}$

Step 3

Solve $\frac{x}{3} + 1 < 1$ when $x \geq - 3$ .

Solve $\frac{x}{3} + 1 < 1$ for $x$ .

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

Subtract $1$ from both sides of the inequality.

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

Subtract $1$ from $1$ .

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

Multiply both sides by $3$ .

$\frac{x}{3} \cdot 3 < 0 \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{x}{3} \cdot 3 < 0 \cdot 3$

Rewrite the expression.

$x < 0 \cdot 3$

Simplify the right side.

Multiply $0$ by $3$ .

$x < 0$

Find the intersection of $x < 0$ and $x \geq - 3$ .

$- 3 \leq x < 0$

Step 4

Solve $- \frac{x}{3} - 1 < 1$ when $x < - 3$ .

Solve $- \frac{x}{3} - 1 < 1$ for $x$ .

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

Add $1$ to both sides of the inequality.

$- \frac{x}{3} < 1 + 1$

Add $1$ and $1$ .

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

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

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

$\frac{- \frac{x}{3}}{- 1} > \frac{2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\frac{x}{3}}{1} > \frac{2}{- 1}$

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

$\frac{x}{3} > \frac{2}{- 1}$

Simplify the right side.

Divide $2$ by $- 1$ .

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

Multiply both sides by $3$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$x > - 2 \cdot 3$

Simplify the right side.

Multiply $- 2$ by $3$ .

$x > - 6$

Find the intersection of $x > - 6$ and $x < - 3$ .

$- 6 < x < - 3$

Step 5

Find the union of the solutions.

$- 6 < x < 0$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$- 6 < x < 0$

Interval Notation:

$(- 6, 0)$

Step 7

Solve the Absolute Value Inequality for x -8|x/3+1|>-8

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