solve that


$\frac{2 x-3}{4 x+5}=\frac{1}{3}$


Given, $\frac{2 x-3}{4 x+5}=\frac{1}{3}$

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

$\Rightarrow \quad 6 x-9=4 x+5$

$\Rightarrow$  $6 x-4 x=9+5 \quad$ [transposing $4 x$ to LHS and $-9$ to RHS]

$\Rightarrow$ $2 x=14$

$\Rightarrow$ $\frac{2 x}{2}=\frac{14}{2}$ [dividing both sides by 2]

$\therefore$ $x=7$

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