Question

The denominator of a fraction is 1 more than double the numerator. On adding 2 to the numerator
and subtracting 3 from the denominator, we obtain 1. Find the original fraction.
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Answer 1

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

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• The denominator of a fraction is 1 more double than the numerator
• on adding 2 to the numerator and subtracting 3 from the denominator we obtain 1

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

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• Original fraction = ??

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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let Numerator be x

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Therefore ; Denominator = 2x + 1

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Acc. to the given statement :-

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$\sf : \implies\: {\bold{ \dfrac{x + 2 }{2x + 1 - 3 } = \dfrac{1}{1} }}$

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$\sf : \implies\: {\bold{ \dfrac{x + 2 }{2x - 2 } = \dfrac{1}{1} }}$

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$\sf : \implies\: {\bold{ 1\times ( x + 2 ) = 1( 2x - 2 ) }}$

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$\sf : \implies\: {\bold{ x + 2 = 2x - 2 }}$

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$\sf : \implies\: {\bold{ 2 + 2 = 2x - x }}$

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$\sf : \implies\: {\bold{ x = 2+2 }}$

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$\sf : \implies\: {\bold{ x = 4 }}$

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• Putting value of x in the fraction

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Numerator = x = 4

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Denominator = 2x + 1 = 2 ×4 + 1 = 8 + 1 = 9

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

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• Original fraction = 4 / 9

Answer 2

Answer:

${ \huge {\bf {\underline {answer}}}}$

Let:-

Numerator = y

Denominator = 2y + 1

Then,

$\sf \: \dfrac{y + 2}{2y + 1 - 3} = \dfrac {1}{1}$

$\sf \dfrac {y + 2}{2y - 2} = \dfrac{1}{1}$

$\sf \: 1(y + 2) =1( 2y - 2)$

$\sf \: {y + 2} = 2y-2$

$\sf \: 2 + 2 = 2y - y$

$\sf \: 4 = y$

$\sf \: denominator \: = 2 \times 4+ 1 = 9$

$\huge \bf \: fraction \: = \frac{4}{9}$