[ Completing the square. Sums of squares ]

[ Substitution of variables (other) $\quad]$

Find all real solutions of the equation

$$
36 \sqrt{x-2}+4 \sqrt{y-1}=28-4 \sqrt{x-2}-\sqrt{y-1}
$$
From the condition, it is clear that $x > 2, y > 1$. Let's introduce the notations: $\sqrt{x-2}=a^{2}, \sqrt{y-1}=b^{2}$. Our equation can be rewritten as: $36 / a^{2}+4 / b^{2}+4 a^{2}+b^{2}-28=0$. We can break the constant term into two addends and regroup the equations:

$$
\left(36 / a^{2}-24+4 a^{2}\right)+\left(4 / b^{2}-4+b^{2}\right)=0
$$

or

$$
(6 / a-2 a)^{2}+(2 / b-b)^{2}=0
$$

But the sum of two squares of real numbers can equal zero only if each addend is zero. Therefore, $6 / a=2 a, 2 / b=b$ or $a^{2}=3, b^{2}=2$. Returning to the original notations, we find: $x=11, y=5$.

## Answer

$x=11, y=5$.