In a fraction, twice the numerator is 2 more than the denominator.

Question:

In a fraction, twice the numerator is 2 more than the denominator. If 3 is added to the numerator and to the denominator, the new fraction is $\frac{2}{3}$. Find the original fraction.

Solution:

Denominator, $\mathrm{d}=\mathrm{x}$

It is given that twice the numerator is equal to two more than the denominator.

$\therefore$ Twice of numerator, $2 \mathrm{n}=\mathrm{x}+2$

$\therefore$ Numerator, $\mathrm{n}=\frac{\mathrm{x}+2}{2}$

$\therefore \frac{n+3}{d+3}=\frac{2}{3}$

$\Rightarrow 3(n+3)=2(d+3) \quad$ (bycross multiplication)

$\Rightarrow 3 n+9=2 d+6$

$\Rightarrow 3 n-2 d=6-9$

$\Rightarrow 3 n-2 d=-3$

On replace $\mathrm{d}$ by $\mathrm{x}$ and $\mathrm{n}$ by $\frac{x+2}{2}$ :

$\Rightarrow 3\left(\frac{x+2}{2}\right)-2 x=-3$

$\Rightarrow \frac{3 x+6-4 x}{2}=-3 \quad($ taking the L. C.M. of 2 and 1 as 2$)$

$\Rightarrow 6-x=-6 \quad$ (by cross multipliction)

$\Rightarrow-x=-6-6$

$\Rightarrow x=12$

The denominator is 12 .

$\therefore$ Numerator $=\frac{x+2}{2}=\frac{12+2}{2}=\frac{14}{2}=7$

$\therefore$ Original fraction $=\frac{7}{12}$

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