prove that


$\frac{9-3 y}{1-9 y}=\frac{8}{5}$


Given, $\frac{9-3 y}{1-9 y}=\frac{8}{5}$

$\Rightarrow$ $5(9-3 y)=8(1-9 y)$ [by cross-multiplication]

$\Rightarrow \quad 45-15 y=8-72 y$

$\Rightarrow \quad 72 y-15 y=8-45 \quad$ [transposing $-72 y$ to LHS and 45 to RHS]

$\Rightarrow \quad 57 y=-37$

$\Rightarrow$ $\frac{57 y}{57}=\frac{-37}{57}$ [dividing both sides by 57 ]

$\therefore$ $y=\frac{-37}{57}$



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