Question

A fraction becomes $\frac{9}{11}$, if $1$ is added to both numerator and denominator. If $3$ is added to both the numerator and the denominator it becomes $\frac{5}{6}$. Find the fraction.

Answer 1

Let the numerator be $x$ and the denominator be $y$.

From the given information, we have,

$\frac{x+1}{y+1}=\frac{9}{11}and\frac{x+3}{y+3}=\frac{5}{6}or,\frac{x+1}{y+1}=\frac{9}{11}=>11(x+1)=9(y+1)=>11x+11=9y+9=>11x-9y+2=0Also,\frac{x+3}{y+3}=\frac{5}{6}=>6(x+3)=5(y+3)=>6x+18=5y+15=>6x-5y+3=0Thus,equationsare:11x-9y+2=06x-5y+3=0$

Using cross-multiplication method,

$\frac{x}{(-9)\left(3\right)-\left(2\right)(-5)}=\frac{y}{\left(2\right)\left(6\right)-\left(3\right)\left(11\right)}=\frac{1}{\left(11\right)\left(-5\right)-\left(6\right)(-9)}\frac{x}{-27+10}=\frac{y}{12-33}=\frac{1}{-55+54}\frac{x}{-17}=\frac{y}{-21}=\frac{1}{-1}=>\frac{x}{-17}=\frac{1}{-1}or,x=17=>\frac{y}{-21}=\frac{1}{-1}or,y=21$

Thus, original fraction is $\frac{x}{y}=\frac{17}{21}$