Question

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Answer 1

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Find x

$\large \rm \: \dfrac{1 - \frac{1 + \frac{1 - x}{3} }{3} }{4} = 1$

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Take LCM

$\rm \mapsto \large \: \frac{1 - \frac{ \frac{3 + 1 - x}{3} }{3} }{4} = 1$

$\large\mapsto \rm \frac{1 - \frac{4 - x}{9} }{4} = 1$

$\large \rm \mapsto \: \frac{9 - (4 - x)}{9} = 1 \times 4$

$\mapsto \rm \: 9 - 4 + x = 4 \times 9$

$\mapsto \rm \: 5 + x = 36$

$\mapsto \rm \: x = 36 - 5$

$\large\mapsto \boxed{ \rm \: x = 31}$

So ,

For x = 31 , this equation is valid ..

★ Verification :

Put x = 31 in LHS

$\rm \large \frac{1 - \frac{1 + \frac{1 - 31}{3} }{3} }{4}$

$\large\implies \rm \frac{1 - \frac{1 + \frac{ - \cancel{30}}{ \cancel3} }{3} }{4}$

$\large \implies \rm \: \frac{1 - \frac{1 - 10}{3} }{4}$

$\large \implies \rm \: \frac{1 - \cancel{ \frac{ - 9}{3} }}{4}$

$\implies \rm \: \frac{1 - (- 3)}{4}$

$\implies \rm \: \frac{1 + 3}{4}$

$\implies \cancel{ \frac{4}{4}}$

$\implies \: 1$

So ,

LHS = RHS , hence verified ..