Solve the Inequality for a -1<=(6-a)/3<=1

$- 1 \leq \frac{6 - a}{3} \leq 1$

Multiply each term in the inequality by $3$ .

$- 1 \cdot 3 \leq \frac{6 - a}{3} \cdot 3 \leq 1 \cdot 3$

Multiply $- 1$ by $3$ .

$- 3 \leq \frac{6 - a}{3} \cdot 3 \leq 1 \cdot 3$

Cancel the common factor of $3$ .

Cancel the common factor.

$- 3 \leq \frac{6 - a}{3} \cdot 3 \leq 1 \cdot 3$

Rewrite the expression.

$- 3 \leq 6 - a \leq 1 \cdot 3$

Multiply $3$ by $1$ .

$- 3 \leq 6 - a \leq 3$

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

Subtract $6$ from each section of the inequality because it does not contain the variable we are trying to solve for.

$- 3 - 6 \leq - a \leq 3 - 6$

Subtract $6$ from $- 3$ .

$- 9 \leq - a \leq 3 - 6$

Subtract $6$ from $3$ .

$- 9 \leq - a \leq - 3$

Multiply each term in the inequality by $- 1$ .

$\frac{- 9}{- 1} \geq \frac{- a}{- 1} \geq \frac{- 3}{- 1}$

Divide $- 9$ by $- 1$ .

$9 \geq \frac{- a}{- 1} \geq \frac{- 3}{- 1}$

Dividing two negative values results in a positive value.

$9 \geq \frac{a}{1} \geq \frac{- 3}{- 1}$

Divide $a$ by $1$ .

$9 \geq a \geq \frac{- 3}{- 1}$

Divide $- 3$ by $- 1$ .

$9 \geq a \geq 3$

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

$3 \leq a \leq 9$

The result can be shown in multiple forms.

Inequality Form:

$3 \leq a \leq 9$

Interval Notation:

$[3, 9]$