Solve the following system of equations:

$$
\begin{array}{r}
x^{2}+y \sqrt{x y}=105 \\
y^{2}+x \sqrt{y x}=70
\end{array}
$$
None of the unknowns is 0 (otherwise at least one of the equations' left sides would be 0), furthermore, they are of the same sign, otherwise $\sqrt{x y}$ would not make sense; let's assume for now that both are positive. Then $\sqrt{x y}=\sqrt{x} \cdot \sqrt{y}=x^{\frac{1}{2}} \cdot y^{\frac{1}{2}}$ and our equations can be transformed as follows:

$$
x^{\frac{1}{2}}\left(x^{\frac{3}{2}}+y^{\frac{3}{2}}\right)=105, \quad y^{\frac{1}{2}}\left(x^{\frac{3}{2}}+y^{\frac{3}{2}}\right)=70
$$

According to the above, the expression in parentheses is positive; by eliminating it (briefly: dividing the two equations), $x^{\frac{1}{2}}: y^{\frac{1}{2}}=105$ : $70=3: 2$, so $y^{\frac{1}{2}}=2 x^{\frac{1}{2}} / 3, y^{\frac{3}{2}}=8 x^{\frac{3}{2}} / 27$, and with this from the first equation $35 x^{2} / 27=105, x^{2}=81, x=9$, and finally $y=4$. This pair of values satisfies both equations.

Another root is not possible, because the $x, y$ with it leads to a contradiction. Indeed, if $x$ and $y$ are negative and $x=y$, then $\sqrt{x y}=\sqrt{x^{2}}=|x|=-x=-y$, and the left side of both equations is 0. If $x \neq y$, from here by multiplying with the negative $x$, $x \sqrt{x y} < 0$. Finally, $y < x^{2}, \sqrt{x y} > -x, y \sqrt{x y} < 0$; based on the assumption that led us to $x^{2}=81$, so $x \neq -9$.