Question

Solve: $\frac{3(y-5)}{4}-4y=3-\frac{(y-3)}{2}$

Answer 1

$\left(\frac{3(y-5)}{4}\right)-4y=3-\left(\frac{(y-3)}{2}\right)$

$\Rightarrow \frac{3(y-5)-16y}{4}=\frac{6-(y-3)}{2}$

$\Rightarrow \frac{3(y-5)-16y}{2}=\frac{6-(y-3)}{1}$

$\Rightarrow 3(y-5)-16y=12-2(y-3)$

$\Rightarrow 3y-15-16y=12-2y+6$

$\Rightarrow 3y-16y+2y=12+6+15$

$\Rightarrow -11y=33$

$\Rightarrow y=-\frac{33}{11}$

$\Rightarrow y=-3$

Hence, the solution of $\left(\frac{3(y-5)}{4}\right)-4y=3-\left(\frac{(y-3)}{2}\right)$ is $y=-3$.