Let the numerator $$ =x $$ and the denominator $$ =y $$
So, the fraction $$ =\dfrac{x}{y} $$
According to the question,
Condition I:
$$ x+y=8 $$
$$ \Rightarrow y=8-x \ldots(i) $$
Condition II:
$$ \dfrac{x+3}{y+3}=\dfrac{3}{4} $$
$$ \Rightarrow 4(x+3)=3(y+3) $$
$$ \Rightarrow 4 x+12=3 y+9 $$
$$ \Rightarrow 4 x-3 y=-3 \ldots$$(ii)
On putting the value of y in Eq.(ii), we get
$$ 4 x-3(8-x)=-3 $$
$$ \Rightarrow 4 x-24+3 x=-3 $$
$$ \Rightarrow 7 x=21 $$
$$ \Rightarrow x=3 $$
On putting the value of x in Eq. (i), we get
$$ y=8-3 $$
$$ \Rightarrow \mathrm{y}=5 $$
So, the numerator is 3 and the denominator is 5
Hence, the fraction is $$ \dfrac{3}{5} $$