Question:
$\frac{2 y-3}{4}-\frac{3 y-5}{2}=y+\frac{3}{4}$
Solution:
Given, $\frac{2 y-3}{4}-\frac{3 y-5}{2}=y+\frac{3}{4}$
$\Rightarrow$ $\frac{2 y-3-2(3 y-5)}{4}=\frac{4 y+3}{4}$
$\Rightarrow \quad 2 y-3-6 y+10=4 y+3$
$\Rightarrow$ $-4 y+7=4 y+3$ [transposing $4 y$ to LHS and 7 to RHS]
$\Rightarrow$ $-4 y-4 y=3-7$
$\Rightarrow$ $-8 y=-4$
$\Rightarrow$ $\frac{-8 y}{-8}=\frac{-4}{-8}$ [dividing both sides by $-8$ ]
$\therefore$ $y=\frac{1}{2}$