Answer
$[5,11]$.
Work Step by Step
The given compound inequality is
$2x+2\geq12$ and $\frac{2x-1}{3}\leq7$.
Solve each inequality separately.
First $2x+2\geq12$.
Subtract $2$ from both sides.
$\Rightarrow 2x+2-2\geq12-2$
Simplify.
$\Rightarrow 2x\geq10$
Divide both sides by $2$.
$\Rightarrow \frac{2x}{2}\geq\frac{10}{2}$
Simplify.
$\Rightarrow x\geq5$
Second $\frac{2x-1}{3}\leq7$.
Multiply both sides by $3$
$\Rightarrow (3)(\frac{2x-1}{3})\leq(3)7$
Simplify.
$\Rightarrow 2x-1\leq21$
Add $1$ to both sides.
$\Rightarrow 2x-1+1\leq21+1$
Simplify.
$\Rightarrow 2x\leq22$
Divide both sides by $2$.
$\Rightarrow \frac{2x}{2}\leq\frac{22}{2}$
Simplify.
$\Rightarrow x\leq11$
First graph then take the intersection of the two inequality.
We can write the compound inequality.
$x\leq11$ as $(-\infty,11]$ and $x\geq5$ as $[5,\infty)$
The intersection is
$(-\infty,11]\cap[5,\infty)=[5,11]$.
