Solve the System of Inequalities 8x>=-24 and 10>2x+6

$8 x \geq - 24$ and $10 > 2 x + 6$

Divide each term in $8 x \geq - 24$ by $8$ and simplify.

Divide each term in $8 x \geq - 24$ by $8$ .

$\frac{8 x}{8} \geq \frac{- 24}{8}$ and $10 > 2 x + 6$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} \geq \frac{- 24}{8}$ and $10 > 2 x + 6$

Divide $x$ by $1$ .

$x \geq \frac{- 24}{8}$ and $10 > 2 x + 6$

Simplify the right side.

Divide $- 24$ by $8$ .

$x \geq - 3$ and $10 > 2 x + 6$

Simplify the second inequality.

Rewrite so $x$ is on the left side of the inequality.

$x \geq - 3$ and $2 x + 6 < 10$

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

Subtract $6$ from both sides of the inequality.

$x \geq - 3$ and $2 x < 10 - 6$

Subtract $6$ from $10$ .

$x \geq - 3$ and $2 x < 4$

Divide each term in $2 x < 4$ by $2$ and simplify.

Divide each term in $2 x < 4$ by $2$ .

$x \geq - 3$ and $\frac{2 x}{2} < \frac{4}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$x \geq - 3$ and $\frac{2 x}{2} < \frac{4}{2}$

Divide $x$ by $1$ .

$x \geq - 3$ and $x < \frac{4}{2}$

Simplify the right side.

Divide $4$ by $2$ .

$x \geq - 3$ and $x < 2$

The intersection consists of the elements that are contained in both intervals.

$- 3 \leq x < 2$

The result can be shown in multiple forms.

Inequality Form:

$- 3 \leq x < 2$

Interval Notation:

$[- 3, 2)$