Solve the System of Inequalities 5x-5>=-5 or -3x+20<=12

$5 x - 5 \geq - 5$ or $- 3 x + 20 \leq 12$

Simplify the first inequality.

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

Add $5$ to both sides of the inequality.

$5 x \geq - 5 + 5$

Add $- 5$ and $5$ .

$5 x \geq 0$

Divide each term in $5 x \geq 0$ by $5$ and simplify.

Divide each term in $5 x \geq 0$ by $5$ .

$\frac{5 x}{5} \geq \frac{0}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} \geq \frac{0}{5}$

Divide $x$ by $1$ .

$x \geq \frac{0}{5}$

Simplify the right side.

Divide $0$ by $5$ .

$x \geq 0$

Simplify the second inequality.

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

Subtract $20$ from both sides of the inequality.

$- 3 x \leq 12 - 20$

Subtract $20$ from $12$ .

$- 3 x \leq - 8$

Divide each term in $- 3 x \leq - 8$ by $- 3$ and simplify.

Divide each term in $- 3 x \leq - 8$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 3 x}{- 3} \geq \frac{- 8}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} \geq \frac{- 8}{- 3}$

Divide $x$ by $1$ .

$x \geq \frac{- 8}{- 3}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x \geq \frac{8}{3}$

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

$x \geq 0$

The result can be shown in multiple forms.

Inequality Form:

$x \geq 0$

Interval Notation:

$[0, \infty)$

Solve the System of Inequalities 5x-5>=-5 or -3x+20<=12

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