Question
Solve the following linear in$-$equation and graph the solution set on a real number line: $4 \frac{3}{4} \geq x +\frac{5}{6}>\frac{1}{3}, x \in R$
✓
Answer
$4 \frac{3}{4} \geq x+\frac{5}{6}$
$\frac{19}{4} \geq \frac{6 x+5}{6}$
$114 \geq 24 x+20$
$114-20 \geq 24 x$
$94 \geq 24 x$
$x \leq 3 \frac{11}{12}$
and
$x+\frac{5}{6}>\frac{1}{3}$
$\frac{6 x+5}{6}>\frac{1}{3}$
$18 x+15>6$
$18 x>6-15$
$18 x>-9$
$x>-\frac{1}{2}$
Solution set $=\left[-\frac{1}{2}< x \leq 3 \frac{11}{12}\right]$
$\frac{19}{4} \geq \frac{6 x+5}{6}$
$114 \geq 24 x+20$
$114-20 \geq 24 x$
$94 \geq 24 x$
$x \leq 3 \frac{11}{12}$
and
$x+\frac{5}{6}>\frac{1}{3}$
$\frac{6 x+5}{6}>\frac{1}{3}$
$18 x+15>6$
$18 x>6-15$
$18 x>-9$
$x>-\frac{1}{2}$
Solution set $=\left[-\frac{1}{2}< x \leq 3 \frac{11}{12}\right]$
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