We write our equations as

$$\begin{aligned} & \sqrt{x}(x \sqrt{x}+y \sqrt{y})=105 \ldots \\ & \sqrt{y}(y \sqrt{y}+x \sqrt{x})=70 \ldots \end{aligned}$$

assuming real factors, $x$ and $y$ can only be positive numbers. Dividing the corresponding sides of (1a) and (2a):

$$\frac{\sqrt{x}}{\sqrt{y}}=\frac{3}{2} \quad \frac{x}{y}=\frac{9}{4} \cdots$$

From (3), $x=\frac{9}{4} y$; substituting this into (2), for example:

$$y^{2}+\frac{9 y}{4} \cdot \frac{3 y}{2}=70, \quad \text { or } \quad 35 y^{2}=8 \cdot 70 \quad \text { and } \quad y^{2}=16$$

Thus,

$$\begin{gathered} y=4 \quad \text { and thus } \quad x=9 \\ y=-4 \quad \text { and } \quad x=-9 \quad \text { satisfy the system of equations } \quad x^{2}-y \sqrt{x y}=105, \quad y^{2}-x \sqrt{x y}=70 \end{gathered}$$

Csáky Gyula (Dobó István g. VI. o. Eger).