Solve the System of Inequalities 3x<-12 or 2x+3>=11

$3 x < - 12$ or $2 x + 3 \geq 11$

Step 1

Divide each term in $3 x < - 12$ by $3$ and simplify.

Divide each term in $3 x < - 12$ by $3$ .

$\frac{3 x}{3} < \frac{- 12}{3}$ or $2 x + 3 \geq 11$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} < \frac{- 12}{3}$ or $2 x + 3 \geq 11$

Divide $x$ by $1$ .

$x < \frac{- 12}{3}$ or $2 x + 3 \geq 11$

Simplify the right side.

Divide $- 12$ by $3$ .

$x < - 4$ or $2 x + 3 \geq 11$

Step 2

Simplify the second inequality.

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

Subtract $3$ from both sides of the inequality.

$x < - 4$ or $2 x \geq 11 - 3$

Subtract $3$ from $11$ .

$x < - 4$ or $2 x \geq 8$

Divide each term in $2 x \geq 8$ by $2$ and simplify.

Divide each term in $2 x \geq 8$ by $2$ .

$x < - 4$ or $\frac{2 x}{2} \geq \frac{8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$x < - 4$ or $\frac{2 x}{2} \geq \frac{8}{2}$

Divide $x$ by $1$ .

$x < - 4$ or $x \geq \frac{8}{2}$

Simplify the right side.

Divide $8$ by $2$ .

$x < - 4$ or $x \geq 4$

Step 3

The result can be shown in multiple forms.

Inequality Form:

$x < - 4 or x \geq 4$

Interval Notation:

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

Step 4