Question

$\frac{1}{3}(x-\frac{1}{2})+\frac{1}{5}(2x+\frac{1}{4})=\frac{3}{4}(x-\frac{1}{12})$

• A. $-8\frac{1}{4}$
• B. $-3\frac{1}{4}$
• C. $-1\frac{1}{4}$
• D. $-7\frac{1}{4}$

Answer 1

Answer: (B)

$-3\frac{1}{4}$

$\frac{1}{3}(x-\frac{1}{2})+\frac{1}{5}(2x+\frac{1}{4})=\frac{3}{4}(x-\frac{1}{12})$
$=>\frac{x}{3}-\frac{1}{6}+\frac{2x}{5}+\frac{1}{20}=\frac{3x}{4}-\frac{3}{48}$
$=>\frac{x}{3}-\frac{1}{6}+\frac{2x}{5}+\frac{1}{20}-\frac{3x}{4}+\frac{3}{48}=0$
$=>\frac{80x-40+96x+12-180x+15}{240}=0$
$=>-4x=13$
$=>x=-\frac{13}{4}$
$=>x=-3\frac{1}{4}$