Solve the System of Inequalities x/-3<=2 or 2x+8<=-10

$\frac{x}{- 3} \leq 2$ or $2 x + 8 \leq - 10$

Simplify the first inequality.

Multiply both sides by $- 3$ .

$\frac{x}{- 3} \cdot - 3 \geq 2 \cdot - 3$

Simplify.

Simplify the left side.

Simplify $\frac{x}{- 3} \cdot - 3$ .

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

$\frac{x}{3 (- 1)} \cdot - 3 \geq 2 \cdot - 3$

Factor $3$ out of $- 3$ .

$\frac{x}{3 \cdot - 1} \cdot (3 \cdot - 1) \geq 2 \cdot - 3$

Cancel the common factor.

$\frac{x}{3 \cdot - 1} \cdot (3 \cdot - 1) \geq 2 \cdot - 3$

Rewrite the expression.

$\frac{x}{- 1} \cdot - 1 \geq 2 \cdot - 3$

Combine $\frac{x}{- 1}$ and $- 1$ .

$\frac{x \cdot - 1}{- 1} \geq 2 \cdot - 3$

Simplify the expression.

Move the negative one from the denominator of $\frac{x \cdot - 1}{- 1}$ .

$- 1 \cdot (x \cdot - 1) \geq 2 \cdot - 3$

Rewrite $- 1 \cdot (x \cdot - 1)$ as $- (x \cdot - 1)$ .

$- (x \cdot - 1) \geq 2 \cdot - 3$

Move $- 1$ to the left of $x$ .

$- (- 1 \cdot x) \geq 2 \cdot - 3$

Rewrite $- 1 x$ as $- x$ .

$- - x \geq 2 \cdot - 3$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$1 x \geq 2 \cdot - 3$

Multiply $x$ by $1$ .

$x \geq 2 \cdot - 3$

Simplify the right side.

Multiply $2$ by $- 3$ .

$x \geq - 6$

Simplify the second inequality.

Move all terms not containing $x$ to the right side of the inequality.

Subtract $8$ from both sides of the inequality.

$2 x \leq - 10 - 8$

Subtract $8$ from $- 10$ .

$2 x \leq - 18$

Divide each term in $2 x \leq - 18$ by $2$ and simplify.

Divide each term in $2 x \leq - 18$ by $2$ .

$\frac{2 x}{2} \leq \frac{- 18}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} \leq \frac{- 18}{2}$

Divide $x$ by $1$ .

$x \leq \frac{- 18}{2}$

Simplify the right side.

Divide $- 18$ by $2$ .

$x \leq - 9$

The union consists of all of the elements that are contained in each interval.

$x \leq - 9$ or $x \geq - 6$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 9 or x \geq - 6$

Interval Notation:

$(- \infty, - 9] \cup [- 6, \infty)$

Solve the System of Inequalities x/-3<=2 or 2x+8<=-10

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