Solve the Absolute Value Inequality for z 3|z+4|-13<=-1

$3 | z + 4 | - 13 \leq - 1$

Step 1

Write $3 | z + 4 | - 13 \leq - 1$ as a piecewise.

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

$z + 4 \geq 0$

Subtract $4$ from both sides of the inequality.

$z \geq - 4$

In the piece where $z + 4$ is non-negative, remove the absolute value.

$3 (z + 4) - 13 \leq - 1$

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

$z + 4 < 0$

Subtract $4$ from both sides of the inequality.

$z < - 4$

In the piece where $z + 4$ is negative, remove the absolute value and multiply by $- 1$ .

$3 (- (z + 4)) - 13 \leq - 1$

Write as a piecewise.

${\begin{cases} 3 (z + 4) - 13 \leq - 1 & z \geq - 4 \\ 3 (- (z + 4)) - 13 \leq - 1 & z < - 4 \end{cases}$

Simplify $3 (z + 4) - 13 \leq - 1$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} 3 z + 3 \cdot 4 - 13 \leq - 1 & z \geq - 4 \\ 3 (- (z + 4)) - 13 \leq - 1 & z < - 4 \end{cases}$

Multiply $3$ by $4$ .

${\begin{cases} 3 z + 12 - 13 \leq - 1 & z \geq - 4 \\ 3 (- (z + 4)) - 13 \leq - 1 & z < - 4 \end{cases}$

Subtract $13$ from $12$ .

${\begin{cases} 3 z - 1 \leq - 1 & z \geq - 4 \\ 3 (- (z + 4)) - 13 \leq - 1 & z < - 4 \end{cases}$

Simplify $3 (- (z + 4)) - 13 \leq - 1$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} 3 z - 1 \leq - 1 & z \geq - 4 \\ 3 (- z - 1 \cdot 4) - 13 \leq - 1 & z < - 4 \end{cases}$

Multiply $- 1$ by $4$ .

${\begin{cases} 3 z - 1 \leq - 1 & z \geq - 4 \\ 3 (- z - 4) - 13 \leq - 1 & z < - 4 \end{cases}$

Apply the distributive property.

${\begin{cases} 3 z - 1 \leq - 1 & z \geq - 4 \\ 3 (- z) + 3 \cdot - 4 - 13 \leq - 1 & z < - 4 \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} 3 z - 1 \leq - 1 & z \geq - 4 \\ - 3 z + 3 \cdot - 4 - 13 \leq - 1 & z < - 4 \end{cases}$

Multiply $3$ by $- 4$ .

${\begin{cases} 3 z - 1 \leq - 1 & z \geq - 4 \\ - 3 z - 12 - 13 \leq - 1 & z < - 4 \end{cases}$

Subtract $13$ from $- 12$ .

${\begin{cases} 3 z - 1 \leq - 1 & z \geq - 4 \\ - 3 z - 25 \leq - 1 & z < - 4 \end{cases}$

Step 2

Solve $3 z - 1 \leq - 1$ when $z \geq - 4$ .

Solve $3 z - 1 \leq - 1$ for $z$ .

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

Add $1$ to both sides of the inequality.

$3 z \leq - 1 + 1$

Add $- 1$ and $1$ .

$3 z \leq 0$

Divide each term in $3 z \leq 0$ by $3$ and simplify.

Divide each term in $3 z \leq 0$ by $3$ .

$\frac{3 z}{3} \leq \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 z}{3} \leq \frac{0}{3}$

Divide $z$ by $1$ .

$z \leq \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$z \leq 0$

Find the intersection of $z \leq 0$ and $z \geq - 4$ .

$- 4 \leq z \leq 0$

Step 3

Solve $- 3 z - 25 \leq - 1$ when $z < - 4$ .

Solve $- 3 z - 25 \leq - 1$ for $z$ .

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

Add $25$ to both sides of the inequality.

$- 3 z \leq - 1 + 25$

Add $- 1$ and $25$ .

$- 3 z \leq 24$

Divide each term in $- 3 z \leq 24$ by $- 3$ and simplify.

Divide each term in $- 3 z \leq 24$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 3 z}{- 3} \geq \frac{24}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 z}{- 3} \geq \frac{24}{- 3}$

Divide $z$ by $1$ .

$z \geq \frac{24}{- 3}$

Simplify the right side.

Divide $24$ by $- 3$ .

$z \geq - 8$

Find the intersection of $z \geq - 8$ and $z < - 4$ .

$- 8 \leq z < - 4$

Step 4

Find the union of the solutions.

$- 8 \leq z \leq 0$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$- 8 \leq z \leq 0$

Interval Notation:

$[- 8, 0]$

Step 6

Solve the Absolute Value Inequality for z 3|z+4|-13<=-1

Related Questions

Support LCMGCF.com by disabling your adblocker.