Solve the Inequality for x 40>=8(12+x)

$40 \geq 8 (12 + x)$

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

$8 (12 + x) \leq 40$

Divide each term in $8 (12 + x) \leq 40$ by $8$ and simplify.

Divide each term in $8 (12 + x) \leq 40$ by $8$ .

$\frac{8 (12 + x)}{8} \leq \frac{40}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 (12 + x)}{8} \leq \frac{40}{8}$

Divide $12 + x$ by $1$ .

$12 + x \leq \frac{40}{8}$

Simplify the right side.

Divide $40$ by $8$ .

$12 + x \leq 5$

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

Subtract $12$ from both sides of the inequality.

$x \leq 5 - 12$

Subtract $12$ from $5$ .

$x \leq - 7$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 7$

Interval Notation:

$(- \infty, - 7]$