Question

A fraction is such that when the denominator is increased by 1 it becomes 1 upon 2 when the numerator is increased by 2 it becomes 1 upon 3 find the fraction please answer​

Answer 1

$\huge\underline{\underline{\bf \bigstar ANSWER \bigstar }}$

Let the numerator be x and the denominator be y .

According to the first condition ,

       $\hookrightarrow\sf \dfrac{x}{y+1} =\dfrac {1}{2} \\\\\hookrightarrow 2x= y+1\\\\\hookrightarrow 2x-y = 1 \ \ \ \ \ \ \ \ \ \ \ ....................(1)$

According to the second condition ,

     $\hookrightarrow \sf \dfrac{x+2}{y}=\dfrac{1}{3}\\\\\hookrightarrow 3(x+2)=y \\\\\hookrightarrow 3x+6 = y \\\\\hookrightarrow 3x-y = -6 \ \ \ \ \ \ \ \ \ ............(2)$

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Eq(1) × 3 ,

  $\longrightarrow\sf 6x-3y=3 \ \ \ \ \ \ \ \ \ \ \ ............(3)$

Eq(2) × 2 ,

  $\longrightarrow \sf 6x-2y = -12 \ \ \ \ \ \ \ \ \ \ \ ............. (4)$

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Eq(3) - Eq(4) ,

    $\Longrightarrow \sf -5y = 15 \\\\\Longrightarrow \ y = -3$

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Put the value of y in Eq(1) ,

    $\hookrightarrow \sf 2x-(3\times(-3)) = 1\\\\\hookrightarrow2x+3 = 1\\\\\hookrightarrow 2x = -2\\\\\hookrightarrow x = -1$

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$\bf FRACTION = \dfrac{-1}{-3}$

HOPE IT HELPS U ............... :)