Example 6. Find the real solutions of the equation
$$
\begin{array}{l}
\frac{36}{\sqrt{2 x-1}}+\frac{4}{\sqrt{3 y+1}} \\
+4 \sqrt{2 x-1}+\sqrt{3 y+1}=28
\end{array}
$$
Let $\sqrt{2^{x-1}}=X$, $\sqrt{3 y+1}=Y$, obviously we have $X>0$, $\boldsymbol{Y}>0$.

Then the original equation can be transformed into
$$
\frac{36}{X}+\frac{4}{Y}+4 X+Y=28 .
$$
(1) can be rewritten as
$$
\begin{array}{l}
\left(2 \sqrt{X}-\frac{6}{2 \bar{X}}\right)^{2} \\
+\left(\sqrt{Y}-\frac{2}{\sqrt{Y}}\right)^{2}=0 .
\end{array}
$$

From (2) we get
$$
2 v \bar{X}-\frac{\beta}{\sqrt{X}}=0
$$

and $\checkmark \bar{Y}-\frac{2}{\sqrt{Y}}=0$.
$$
\therefore \quad X=3, \quad Y=2 \text {. }
$$

Thus, $\sqrt{2 x-1}=3, \sqrt{3 y+1}=2$.
$$
\therefore x=5, \quad y=1 \text {. }
$$

Therefore, the real solutions to the original equation are
$$
\left\{\begin{array}{l}
x=5, \\
y=1 .
\end{array}\right.
$$