Question

The numerator of a fraction is 2 less than the denominator. If 3 is added to both the numerator
and the denominator, the fraction becomes ¾. Find the fraction.

Answer 1

$\huge{\underline{\purple{\rm{Solution:}}}}$

$\large{\underline{\red{\rm{Let:}}}}$

• $\sf{The\: numerator\:be:R}$
• $\sf{The\: denominator\:be:S}$

$\large{\underline{\red{\rm{To\: Find:}}}}$

• $\sf{The\: fraction\:=?}$

$\small{\underline{\purple{\rm{As\:per\:the\: given\: Question:}}}}$

$\sf{The\:N\:is\:2\:less\:than\: it's\:D}$

$\rm{\implies R=S-2}$

$\rm\green{\implies R=S-2-----(1)}$

$\sf{If\:3\:is\:added\:to\:both\:N\:and\:D\:the\: fraction\: become\:\frac{3}{4}}$

$\rm{\implies \dfrac{R+3}{S+3}=\dfrac{3}{4}}$

$\rm{\implies 4(R+3)=3(S+3)}$

$\rm{\implies 4R+12=3S+9}$

$\rm{\implies 4R-3S=9-12}$

$\rm\green{\implies 4R-3S=-3-----(2)}$

• $\sf{putting\:the\: value\:of\:S\:in\:eq\:2}$

$\rm{\implies 4R-3S=-3}$

$\rm{\implies 4(S-2)-3S=-3}$

$\rm{\implies 4S-8-3S=-3}$

$\rm{\implies 4S-3S=-3+8}$

$\rm\red{\implies S=5}$

• $\sf{putting\:the\: value\:of\:S\:in\:eq\:1}$

$\rm{\implies R=S-2}$

$\rm{\implies R=5-2}$

$\rm\red{\implies R=3}$

Hence,

$\rm\green{\implies Required_{fraction}=\dfrac{R}{S}=\dfrac{3}{5}}$

Answer 2

Given:

The numerator of the fraction is 2 less than the denominator.

To be found:

The fraction.

Solution:

Let the denominator be x

Now according to the question

Numerator = 2 less than the denominator

Numerator = (x - 2)

The fraction becomes =  $\bf \frac{(x-2)}{x}$.

Also, If we add 3 to both the numerator and denominator than the fraction becomes = $\bf \frac{3}{4}$.

So,

$\rightarrow \bf \frac{(x-2)+3}{x+3}=\frac{3}{4} }$

$\rightarrow \bf \frac{x+1}{x+3}=\frac{3}{4}$

$\rightarrow \bf 3(x+3)=4(x+1)$

$\therefore \tt{cross\:\:multiplication}$

$\rightarrow \bf 3x+9=4x+4$

$\rightarrow \bf 9-4=4x-3x$

$\therefore \tt Taking\:\: 4 \:\:to \:\: LHS \:\:side\:\:and \:\:3x \:\:to\:\:RHS\:\;side$

$\rightarrow \boxed{\boxed{\bf 5=x}}$

We got the denominator = x = 5

So, the numerator = 2 less than the denominator = (x - 2) = (5 - 2) = 3

Therefore,

$\boxed{\boxed{\bf Required \:\:fraction= \frac{3}{5}}}$

Verification

$\tt adding \:\:3\:\: to\:\: both\:\:numerator\:\:and\:\:denominator$

$= \bf \frac{3+3}{5+3} = \frac{6}{8} =\frac{3}{4} \\ \\ \therefore \frac{6}{2}=3 \:\:and \:\:\frac{8}{2}=4$

Hence, Verified