## Algebra 2 Common Core

$-\frac{1}{2}$ $\leq$ $d$ $\leq$ $\frac{25}{4}$
$3|d-4$ $\leq$ $13-d$ Divide both sides by $3$ $|d-4|$ $\leq$ $\frac{13}{3}-\frac{1}{3}d$ Rewrite as a compound inequality $d-4$ $\leq$ $\frac{13}{3}-\frac{1}{3}d$ or $d-4$ $\geq$ $-\frac{13}{3}+\frac{1}{3}d$ Work through the first equation $d-4$ $\leq$ $\frac{13}{3}-\frac{1}{3}d$ Add $\frac{1}{3}d$ to both sides $\frac{4}{3}d-4$ $\leq$ $\frac{13}{3}$ Add $4$ to each side $\frac{4}{3}d$ $\leq$ $\frac{25}{3}$ Multiply each side by $\frac{3}{4}$ $d$ $\leq$ $\frac{25}{4}$ Now work through the second equation $d-4$ $\geq$ $-\frac{13}{3}+\frac{1}{3}d$ Subtract $\frac{1}{3}d$ from both sides $\frac{2}{3}d-4$ $\geq$ $-\frac{13}{3}$ Add $4$ to both sides $\frac{2}{3}d$ $\geq$ $-\frac{1}{3}$ Multiply each side by $\frac{3}{2}$ $d$ $\geq$ $-\frac{1}{2}$ $-\frac{1}{2}$ $\leq$ $d$ $\leq$ $\frac{25}{4}$