Question 66

Solve the following:

1-(x-2)-[(x-3)-(x-1)]=0

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Solution

Given, 1-(x-2)-[(x-3)-(x-1)]=0
$\Rightarrow 1 - x + 2 - [x - 3 - x + 1] = 0 \Rightarrow 3 - x - [- 2] = 0 \Rightarrow 3 - x + 2 = 0 \Rightarrow 5 - x = 0 \Rightarrow - x = - 5 \Rightarrow \frac{- x}{- 1} = \frac{- 5}{- 1} ∴ x = 5$

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