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prove that

Question:

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

Solution:

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

$\Rightarrow$ $\frac{9 x-(x-2)}{3}=\frac{16-(x-1)}{4}$

$\Rightarrow$ $4(9 x-x+2)=3(16-x+1)$ [by cross-multiplication]

$\Rightarrow$ $4(8 x+2)=3(-x+17)$

$\Rightarrow$ $32 x+3 x=51-8 \quad$ [transposing $-3 x$ to LHS and 8 to RHS]

$\Rightarrow$ $35 x=43$

$\Rightarrow$ $\frac{35 x}{35}=\frac{43}{35}$ [dividing both sides by 35 ]

$\therefore$ $x=\frac{43}{35}$

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