If $2$ is added to both numerator and the denominator then the fraction becomes $9/11$. If $3$ is added to both the numerator and the denominator it becomes $5/6$. Find the fraction.

Let’s assume the numerator of the fraction to be A and the denominator of the fraction to be B.

So, the required fraction is $A/B$.

ATQ , 

the equation so formed is,

$A+2B+2=9/11$

⇒ $11\left( A+2 \right)=9\left( B+2 \right)$

⇒ $11A+22=9B+18$

⇒ $11A-9B=18-22$

⇒ $11A–9B+4=0$ ……. (i)

So ATQ, 

We have, 

$A+3B+3=56$

⇒ $6\left( A+3 \right)=5\left( B+3 \right)$

⇒$6A+18=5B+15$

⇒ $6A-5B=15-18$

⇒ $6A–5B+3=0$…….. (ii)

Solving (i) and (ii), to find the fraction

By using cross-multiplication method, we have

$A-9\times 3-5\times 4=-B11\times 3-6\times 4=111\times -5-6\times -9$

$A-27+20=-B33-24=1-55+54$

$A-7=-B9=1-1$

$A7=B9=1$

 $A=7,B=9$

Hence, the required fraction is $7/9$.