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 $7/9$.