Question

Evaluate:

$\frac{1}{2}\left(\frac{2y+3}{3}\right)+\frac{1}{2}\left(\frac{17-3y}{52}\right)=-3y-\frac{5}{2}+\frac{2y+14}{5}$

Answer 1

Given equation is,

$\frac{1}{2}\left(\frac{2y+3}{3}\right)+\frac{1}{2}\left(\frac{17-3y}{52}\right)=-3y-\frac{5}{2}+\frac{2y+14}{5}$

$\Rightarrow \frac{1}{2}\{\left(\frac{2y+3}{3}\right)+\left(\frac{17-3y}{52}\right)\}=-3y-\frac{5}{2}+\frac{2y+14}{5}$

$\Rightarrow \left(\frac{2y+3}{3}\right)+\left(\frac{17-3y}{52}\right)=-6y-\frac{5\times 2}{2}+\frac{2y+14}{5}\times 2$

$\Rightarrow \frac{(2y+3)52+(17-3y)3}{156}=-6y-5+\frac{4y+28}{5}$

$\Rightarrow \frac{104y+156+51-9y}{156}=-6y-5+\frac{4y+28}{5}$

$\Rightarrow \frac{95y}{156}+\frac{207}{156}=-6y-5+\frac{4y}{5}+\frac{28}{5}$

$\Rightarrow \frac{95y}{156}+6y-\frac{4y}{5}=-5+\frac{28}{5}-\frac{207}{156}$

$\Rightarrow \frac{95y+936y}{156}-\frac{4y}{5}=\frac{-25+28}{5}-\frac{207}{156}$

$\Rightarrow \frac{1031y}{156}-\frac{4y}{5}=\frac{3}{5}-\frac{207}{156}$

$\Rightarrow \frac{5155‬y-624y}{156\times 5}=\frac{468-1035}{5\times 156}$

$\Rightarrow \frac{4531‬y}{156\times 5}=\frac{-567}{5\times 156}$

$\Rightarrow 4531‬y=-567$

$\therefore y=-\frac{567}{4531}$