Solve for z -15<=-3/2z-3<=3

$- 15 \leq - \frac{3}{2} z - 3 \leq 3$

Step 1

Simplify each term.

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

$- 15 \leq - \frac{z \cdot 3}{2} - 3 \leq 3$

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

$- 15 \leq - \frac{3 z}{2} - 3 \leq 3$

Step 2

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

Add $3$ to each section of the inequality because it does not contain the variable we are trying to solve for.

$- 15 + 3 \leq - \frac{3 z}{2} \leq 3 + 3$

Add $- 15$ and $3$ .

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

Multiply each term in the inequality by $2$ .

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

Multiply $- 12$ by $2$ .

$- 24 \leq - \frac{3 z}{2} \cdot 2 \leq 3 \cdot 2 + 3 \cdot 2$

Cancel the common factor of $2$ .

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

$- 24 \leq \frac{- 3 z}{2} \cdot 2 \leq 3 \cdot 2 + 3 \cdot 2$

Cancel the common factor.

$- 24 \leq \frac{- 3 z}{2} \cdot 2 \leq 3 \cdot 2 + 3 \cdot 2$

Rewrite the expression.

$- 24 \leq - 3 z \leq 3 \cdot 2 + 3 \cdot 2$

Simplify each term.

Multiply $3$ by $2$ .

$- 24 \leq - 3 z \leq 6 + 3 \cdot 2$

Multiply $3$ by $2$ .

$- 24 \leq - 3 z \leq 6 + 6$

Add $6$ and $6$ .

$- 24 \leq - 3 z \leq 12$

Step 3

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

$\frac{- 24}{- 3} \geq \frac{- 3 z}{- 3} \geq \frac{12}{- 3}$

Step 4

Divide $- 24$ by $- 3$ .

$8 \geq \frac{- 3 z}{- 3} \geq \frac{12}{- 3}$

Step 5

Cancel the common factor of $- 3$ .

Cancel the common factor.

$8 \geq \frac{- 3 z}{- 3} \geq \frac{12}{- 3}$

Divide $z$ by $1$ .

$8 \geq z \geq \frac{12}{- 3}$

Step 6

Divide $12$ by $- 3$ .

$8 \geq z \geq - 4$

Step 7

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

$- 4 \leq z \leq 8$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$- 4 \leq z \leq 8$

Interval Notation:

$[- 4, 8]$

Step 9

Solve for z -15<=-3/2z-3<=3

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