Solve: $\frac{x-(7-8 x)}{9 x-(3+4 x)}=\frac{2}{3}$.

Given: $\frac{x-(7-8x)}{9 x-(3+4x)}=\frac{2}{3}$.

To do:  To solve: $\frac{x-(7-8x)}{9 x-(3+4x)}=\frac{2}{3}$.

Solution:

$\frac{x-(7-8x)}{9 x-(3+4x)}=\frac{2}{3}$

$\Rightarrow \frac{x-7+8x}{9 x-3-4x}=\frac{2}{3}$

$\Rightarrow \frac{9x-7}{5x-3}=\frac{2}{3}$

$\Rightarrow 3( 9x-7)=2( 5x-3)$

$\Rightarrow 27x-21=10x-6$

$\Rightarrow 27x-10x=-6+21$

$\Rightarrow 17x=15$

$\Rightarrow x=\frac{15}{17}$

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Updated on: 10-Oct-2022

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