Problem 11.3. The product of positive numbers $a$ and $b$ is 1. It is known that

$$
(3 a+2 b)(3 b+2 a)=295
$$

Find $a+b$.
Answer: 7.

Solution. Expanding the brackets, we get

$$
295=6 a^{2}+6 b^{2}+13 a b=6\left(a^{2}+b^{2}\right)+13
$$

from which $a^{2}+b^{2}=47$. Then

$$
(a+b)^{2}=a^{2}+b^{2}+2 a b=47+2=49=7^{2}
$$

which gives $a+b=7$ (note that $a+b>0$, since $a>0$ and $b>0$).