Solve for z -10<=-3/2z-1<=-4

$- 10 \leq - \frac{3}{2} z - 1 \leq - 4$

Step 1

Simplify each term.

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

$- 10 \leq - \frac{z \cdot 3}{2} - 1 \leq - 4$

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

$- 10 \leq - \frac{3 z}{2} - 1 \leq - 4$

Step 2

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

Add $1$ to each section of the inequality because it does not contain the variable we are trying to solve for.

$- 10 + 1 \leq - \frac{3 z}{2} \leq - 4 + 1$

Add $- 10$ and $1$ .

$- 9 \leq - \frac{3 z}{2} \leq - 4 + 1$

Multiply each term in the inequality by $2$ .

$- 9 \cdot 2 \leq - \frac{3 z}{2} \cdot 2 \leq - 4 \cdot 2 + 1 \cdot 2$

Multiply $- 9$ by $2$ .

$- 18 \leq - \frac{3 z}{2} \cdot 2 \leq - 4 \cdot 2 + 1 \cdot 2$

Cancel the common factor of $2$ .

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

$- 18 \leq \frac{- 3 z}{2} \cdot 2 \leq - 4 \cdot 2 + 1 \cdot 2$

Cancel the common factor.

$- 18 \leq \frac{- 3 z}{2} \cdot 2 \leq - 4 \cdot 2 + 1 \cdot 2$

Rewrite the expression.

$- 18 \leq - 3 z \leq - 4 \cdot 2 + 1 \cdot 2$

Simplify each term.

Multiply $- 4$ by $2$ .

$- 18 \leq - 3 z \leq - 8 + 1 \cdot 2$

Multiply $2$ by $1$ .

$- 18 \leq - 3 z \leq - 8 + 2$

Add $- 8$ and $2$ .

$- 18 \leq - 3 z \leq - 6$

Step 3

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

$\frac{- 18}{- 3} \geq \frac{- 3 z}{- 3} \geq \frac{- 6}{- 3}$

Step 4

Divide $- 18$ by $- 3$ .

$6 \geq \frac{- 3 z}{- 3} \geq \frac{- 6}{- 3}$

Step 5

Cancel the common factor of $- 3$ .

Cancel the common factor.

$6 \geq \frac{- 3 z}{- 3} \geq \frac{- 6}{- 3}$

Divide $z$ by $1$ .

$6 \geq z \geq \frac{- 6}{- 3}$

Step 6

Divide $- 6$ by $- 3$ .

$6 \geq z \geq 2$

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.

$2 \leq z \leq 6$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$2 \leq z \leq 6$

Interval Notation:

$[2, 6]$

Step 9

Solve for z -10<=-3/2z-1<=-4

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