Question:
1-(x-2)-[(x-3)-(x-1)]=0
Solution:
Given, $\quad 1-(x-2)-[(x-3)-(x-1)]=0$
$\Rightarrow$ $1-x+2-[x-3-x+1]=0$
$\Rightarrow \quad 3-x-[-2]=0$
$\Rightarrow \quad 3-x+2=0$
$\Rightarrow \quad-x+5=0$
$\Rightarrow$ $-x+5=0$
$\Rightarrow$ $-x=-5$ [transposing 5 to RHS]
$\Rightarrow$ $\frac{-x}{-1}=\frac{-5}{-1}$ [dividing both sides by -1]
$\therefore$ $x=5$
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