prove that


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


Given, $\frac{y-(4-3 y)}{2 y-(3+4 y)}=\frac{1}{5}$

$\Rightarrow$ $5(y-4+3 y)=2 y-3-4 y$ [by cross-multiplication]

$\Rightarrow \quad 5(4 y-4)=-3-2 y$

$\Rightarrow$ $20 y-20=-3-2 y$

$\Rightarrow \quad 20 y+2 y=20-3[$ transposing $-20$ to RHS and $-2 y$ to LHS $]$

$\Rightarrow \quad 22 y=17$

$\Rightarrow$ $\frac{22 y}{22}=\frac{17}{22}$ [dividing both sides by 22 ]

$\therefore$ $y=\frac{17}{22}$

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