Question

A fraction becomes $\frac{9}{11}$ if $2$ is added to both the numerator and the denominator. If $3$ is added to both the numerator and the denominator, it becomes $\frac{5}{6}$. Find the fraction.

Answer 1

Let the fraction be $\frac{x}{y}$, $y\ne 0$

Case1:

fraction becomes $\frac{9}{11}$ if $2$ is added to both the numerator and the denominator.

$\Rightarrow \frac{x+2}{y+2}=\frac{9}{11}$

$\Rightarrow 11(x+2)=9(y+2)$

$\Rightarrow 11x+22=9y+18$

$\Rightarrow 11x-9y+22-18=0$

$\Rightarrow 11x-9y+4=0$ ---(1)

Case2:

fraction becomes $\frac{5}{6}$ if $3$ is added to both the numerator and the denominator.

$\Rightarrow \frac{x+3}{y+3}=\frac{5}{6}$

$\Rightarrow 6(x+3)=5(y+3)$

$\Rightarrow 6x+18=5y+15$

$\Rightarrow 6x-5y+18-15=0$

$\Rightarrow 16x-5y+3=0$ ---(2)

Lets solve (1) and (2)

Multiply $11x-9y+4=0$ by $5$

$\Rightarrow 55x-45y+20=0$----(3)

Multiply $6x-5y+3=0$ by $9$

$\Rightarrow 54x-45y+27=0$ ---(4)

Lets do (3)-(4)

$\Rightarrow 55x-45y+20-(54x-45y+27)=0$

$\Rightarrow 55x-45y+20-54x+45y-27)=0$

$\Rightarrow x-7=0$

$\Rightarrow x=7$

Substitute $x=7$ in $11x-9y+4=0$

$\Rightarrow 11\left(7\right)-9y+4=0$

$\Rightarrow 77-9y+4=0$

$\Rightarrow 81-9y=0$

$\Rightarrow 81=9y$

$\Rightarrow \frac{81}{9}=y$

$\Rightarrow 9=y$

$\therefore$ The required fraction is $\frac{7}{9}$.