If positive reals $a$ and $b$ satisfy $a\sqrt{a}+b\sqrt{b}=183, a\sqrt{b}+b\sqrt{a}=182$, find $\frac{9}{5}(a+b)$. Question from Math Olympiad:

Suppose $a, b$ are positive real numbers such that $a\sqrt{a} + b\sqrt {b} = 183$ and $a\sqrt{b} + b\sqrt {a} = 182$. Find $\frac{9}{5}(a+b)$.

My approach:
$a\sqrt{a} + b\sqrt {b} = 183$
$a\sqrt{b} + b\sqrt {a} = 182$
Therefore $(\sqrt{a} + \sqrt{b})(a+b) = 365$
I only see two integral possibilities:
$1 \times 365 $  and $5 \times 73 $
And on putting either of them, I am getting $657$ and $131.4$, whereas the answer is an integral $73$.
Please help me with the general method to go about the same.

This is the picture from the source.

Hint : If you make the difference of the two equations, you get
$$(\sqrt{a}-\sqrt{b})(a-b)=0$$
Probably you can conclude.
