Question

It is a question of linear equations. Pls Help it is urgent​

Answer 1

QUESTION:-

$\bf\;\frac{3(y-5)}{4}-4y=3-\frac{y-3}{2}$

EXPLANATION:-

Simplifying the linear equation we get:-

$\bf\;\frac{3(y-5)}{4}-4y=3-\frac{y-3}{2}\\\\\\=>\frac{3(y-5)-16y}{4}=\frac{6-y-3 }{2}\;\;\;\;[TAKING\;LCM]\\\\\\=>4(6-y-3)=2[3(y-5)-5]\;\;[CROSS\;MULTIPLICATION]\\\\\\=>24-4y-12=2(3y-15-5)\\\\\\=>12-4y=6y-30-10\\\\\\=>12-4y=6y-40\\\\\\=>-4y+6y=-40-12\\\\\\=>2y=-52\\\\\\=>y=\frac{-52}{2}\\\\\\=>\underline{ y=-26}$

So the value of y  is -26

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