Solving the system $x\sqrt{y} + y\sqrt{x} = 30$, $x\sqrt{x} + y\sqrt{y} = 35$

algebra-precalculusradicalssystems of equations

I'm stuck on this problem:
$$ \begin{cases} x\sqrt{y} + y\sqrt{x} = 30 \\ x\sqrt{x} + y\sqrt{y} = 35\end{cases} $$

I've not solved this kind of problem before. I tried formula for square of sum and also sum of cubes to possibly isolate an expression. I've also tried substituting $ \sqrt{xy}$ or some other expression with a new variable, but couldn't really get anywhere.

P.S. I'm aware that you can easily get the answers by just trying to plug in the numbers, I'm looking for a more general, algebraic solution.

Best Answer

From

$$\begin{cases} x\sqrt{y} + y\sqrt{x} = 30 & \ \ (a)\\ x\sqrt{x} + y\sqrt{y} = 35& \ \ (b)\end{cases}$$

$3 \times$(a) + (b) gives :

$$(\sqrt{x}+\sqrt{y})^3=5^3\ \ \iff \ \ \sqrt{x}+\sqrt{y}=5 \ \tag{1}$$

Besides, (b) - (a) gives :

$$x(\sqrt{x}-\sqrt{y})-y(\sqrt{x}-\sqrt{y})=5 \ \iff$$

$$(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})=5 \tag{2}$$

Taking (1) into account in (2), on gets :

$$\sqrt{x}-\sqrt{y}=\pm1\tag{3}$$

Gathering (1) and (3), we obtain easily the two solutions :

$$(x,y)=(4,9) \ \text{and} \ (x,y)=(9,4) $$

as confirmed by the following graphical representation ((a) in blue and (b) in red) :

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