Find $x$, if $\sqrt{x} + 2y^2 = 15$ and $\sqrt{4x} − 4y^2 = 6$.

We have
\[\sqrt{4x} ? 4y^2 = 6 \ \ \Rightarrow \ \ 2\sqrt{x}-4y^2=6  \Rightarrow \sqrt{x}-2y^2=3.\]
So the two equations can be written as
\[(1)         \sqrt{x}+2y^2=15, \ \ \sqrt{x}-2y^2=3.\]
Multiplying these two equaitions together we get 
\[(\sqrt{x}+2y^2)(\sqrt{x}-2y^2)=45 \Rightarrow x-4y^4=45 \Rightarrow x=4y^4+45.\]
Substituting $x=4y^4+45$ in the second equation in (1) we get
\[\sqrt{4y^4+45}-2y^2=3 \Rightarrow \sqrt{4y^4+45}=2y^2+3 \Rightarrow 4y^4+45=(2y^2+3)^2.\]
Hence
\[4y^4+45=4y^4+12y^2+9 \Rightarrow 12y^2=36 \Rightarrow y^2=3.\]
So we have
\[x=4y^4+45=4(y^2)^2+45=4(3)^2+45=36+45=81.\]