Solve for z -8<=-3/2z-2<=4

$- 8 \leq - \frac{3}{2} z - 2 \leq 4$

Simplify each term.

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

$- 8 \leq - \frac{z \cdot 3}{2} - 2 \leq 4$

Move $3$ to the left of $z$ .

$- 8 \leq - \frac{3 z}{2} - 2 \leq 4$

Move all terms not containing $z$ from the center section of the inequality.

Add $2$ to each section of the inequality because it does not contain the variable we are trying to solve for.

$- 8 + 2 \leq - \frac{3 z}{2} \leq 4 + 2$

Add $- 8$ and $2$ .

$- 6 \leq - \frac{3 z}{2} \leq 4 + 2$

Multiply each term in the inequality by $2$ .

$- 6 \cdot 2 \leq - \frac{3 z}{2} \cdot 2 \leq 4 \cdot 2 + 2 \cdot 2$

Multiply $- 6$ by $2$ .

$- 12 \leq - \frac{3 z}{2} \cdot 2 \leq 4 \cdot 2 + 2 \cdot 2$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3 z}{2}$ into the numerator.

$- 12 \leq \frac{- 3 z}{2} \cdot 2 \leq 4 \cdot 2 + 2 \cdot 2$

Cancel the common factor.

$- 12 \leq \frac{- 3 z}{2} \cdot 2 \leq 4 \cdot 2 + 2 \cdot 2$

Rewrite the expression.

$- 12 \leq - 3 z \leq 4 \cdot 2 + 2 \cdot 2$

Simplify each term.

Multiply $4$ by $2$ .

$- 12 \leq - 3 z \leq 8 + 2 \cdot 2$

Multiply $2$ by $2$ .

$- 12 \leq - 3 z \leq 8 + 4$

Add $8$ and $4$ .

$- 12 \leq - 3 z \leq 12$

Divide each term in the inequality by $- 3$ .

$\frac{- 12}{- 3} \geq \frac{- 3 z}{- 3} \geq \frac{12}{- 3}$

Divide $- 12$ by $- 3$ .

$4 \geq \frac{- 3 z}{- 3} \geq \frac{12}{- 3}$

Cancel the common factor of $- 3$ .

Cancel the common factor.

$4 \geq \frac{- 3 z}{- 3} \geq \frac{12}{- 3}$

Divide $z$ by $1$ .

$4 \geq z \geq \frac{12}{- 3}$

Divide $12$ by $- 3$ .

$4 \geq z \geq - 4$

Rewrite the interval so that the left value is less than the right value. This is the correct way to write an interval solution.

$- 4 \leq z \leq 4$

The result can be shown in multiple forms.

Inequality Form:

$- 4 \leq z \leq 4$

Interval Notation:

$[- 4, 4]$

Solve for z -8<=-3/2z-2<=4

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