Solve the Inequality for x -5<=(4-3x)/2<1

$- 5 \leq \frac{4 - 3 x}{2} < 1$

Multiply each term in the inequality by $2$ .

$- 5 \cdot 2 \leq \frac{4 - 3 x}{2} \cdot 2 < 1 \cdot 2$

Multiply $- 5$ by $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$- 10 \leq 4 - 3 x < 1 \cdot 2$

Multiply $2$ by $1$ .

$- 10 \leq 4 - 3 x < 2$

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

Subtract $4$ from each section of the inequality because it does not contain the variable we are trying to solve for.

$- 10 - 4 \leq - 3 x < 2 - 4$

Subtract $4$ from $- 10$ .

$- 14 \leq - 3 x < 2 - 4$

Subtract $4$ from $2$ .

$- 14 \leq - 3 x < - 2$

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

$\frac{- 14}{- 3} \geq \frac{- 3 x}{- 3} > \frac{- 2}{- 3}$

Dividing two negative values results in a positive value.

$\frac{14}{3} \geq \frac{- 3 x}{- 3} > \frac{- 2}{- 3}$

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{14}{3} \geq \frac{- 3 x}{- 3} > \frac{- 2}{- 3}$

Divide $x$ by $1$ .

$\frac{14}{3} \geq x > \frac{- 2}{- 3}$

Dividing two negative values results in a positive value.

$\frac{14}{3} \geq x > \frac{2}{3}$

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

$\frac{2}{3} < x \leq \frac{14}{3}$

The result can be shown in multiple forms.

Inequality Form:

$\frac{2}{3} < x \leq \frac{14}{3}$

Interval Notation:

$(\frac{2}{3}, \frac{14}{3}]$