solve for b. $\frac{b + 6}{3}<2$ or $b + 11geq18$ write your answer as …

$solve for b. $\frac{b + 6}{3}<2$ or $b + 11geq18$ write your answer as …$

solve for b.

$\frac{b + 6}{3}<2$ or $b + 11geq18$

write your answer as a compound inequality with integers.

Multiply both sides of $\frac{b + 6}{3}<2$ by 3: $b + 6<6$. Then subtract 6 from both sides: $b<0$.

Subtract 11 from both sides of $b + 11\geq18$: $b\geq7$.

$b<0$ or $b\geq7$