a(a+b)=2418, b(a+b)=3666, then b-a=?

$$\begin{gathered} a\left(a+b\right)=2418......\left(i\right) \\ b\left(a+b\right)=3666.....\left(ii\right) \end{gathered}$$

Now dividing (ii) by (i) we get;
$$\begin{gathered} \frac{b\left(a+b\right)}{a\left(a+b\right)}=\frac{3666}{2418} \\ \Rightarrow \frac{b}{a}=\frac{611}{403} \\ \Rightarrow b=\frac{611}{403}a....\left(iii\right) \end{gathered}$$
Now considering (i) we have;
$$\begin{gathered} a^2+ab=2418 \\ \Rightarrow a^2+a\left(\frac{611}{403}\right)a=2418 \end{gathered}$$  ( taking the value of b from i)
$$\begin{gathered} a^2+\left(\frac{611}{403}\right)a^2=2418 \\ \Rightarrow \frac{403a^2+611a^2}{403}=2418 \\ \Rightarrow \frac{1014a^2}{403}=2418 \\ \Rightarrow a^2=\frac{2418\times 403}{1014}=\frac{403\times 403}{169} \\ \Rightarrow a=\sqrt{\frac{403\times 403}{13\times 13}}=\frac{403}{13}=31 \end{gathered}$$
Putting the value of a in (iii) we get;
$$\begin{gathered} b=\frac{611}{403}\times 31=\frac{611}{13} \\ \Rightarrow b=47 \end{gathered}$$

So $b-a=47-31=16$