Solve the System of Inequalities 4x<=-16 or 3x+5>=17

$4 x \leq - 16$ or $3 x + 5 \geq 17$

Divide each term in $4 x \leq - 16$ by $4$ and simplify.

Divide each term in $4 x \leq - 16$ by $4$ .

$\frac{4 x}{4} \leq \frac{- 16}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} \leq \frac{- 16}{4}$

Divide $x$ by $1$ .

$x \leq \frac{- 16}{4}$

Simplify the right side.

Divide $- 16$ by $4$ .

$x \leq - 4$

Simplify the second inequality.

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

Subtract $5$ from both sides of the inequality.

$3 x \geq 17 - 5$

Subtract $5$ from $17$ .

$3 x \geq 12$

Divide each term in $3 x \geq 12$ by $3$ and simplify.

Divide each term in $3 x \geq 12$ by $3$ .

$\frac{3 x}{3} \geq \frac{12}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{12}{3}$

Divide $x$ by $1$ .

$x \geq \frac{12}{3}$

Simplify the right side.

Divide $12$ by $3$ .

$x \geq 4$

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

$x \leq - 4$ or $x \geq 4$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 4 or x \geq 4$

Interval Notation:

$(- \infty, - 4] \cup [4, \infty)$