$x\sqrt{y}-y\sqrt{z}=42\sqrt{y}$ $y\sqrt{z}-z\sqrt{x}=6\sqrt{z}$ $z\sqrt{x}-x\sqrt{y}=30\sqrt{x}$ Adding member to member $x\sqrt{y}-y\sqrt{z}+y\sqrt{z}-z\sqrt{x}+z\sqrt{x}-x\sqrt{y}=30\sqrt{x}+42\sqrt{y}+6\sqrt{z}$ $5\sqrt{x}+7\sqrt{y}+\sqrt{z}=0\Rightarrow \sqrt{z}=-\left( 5\sqrt{x}+7\sqrt{y} \right)$ Note that $sqrt(z)$ is positive for every real number z, then it could never be a negative number, which implies that there is no z in the real numbers. Similarly we could deduce from $sqrt (x)$ and sqrt $(y)$ $x+\sqrt{y}\cdot \left( 5\sqrt{x}+7\sqrt{y} \right)=42$ $y+\sqrt{x}\left( 5\sqrt{x}+7\sqrt{y} \right)=6$ $\left( 5\sqrt{x}+7\sqrt{y} \right)^{2}-\sqrt{xy}=30$ If we subtract the equation (1) -7x (6) we obtain $0=34x+44\sqrt{xy}=\sqrt{x}\left( \underbrace{34\sqrt{x}+44\sqrt{y}}_{>0} \right)\Rightarrow x=0\Rightarrow y=\frac{30}{49}\Rightarrow z=30$ Which leads to false solutions