Solve $xy=3$ and $4^{x^2}+2^{y^2}=72$ I have a system of equations $xy=3$ and $4^{x^2}+2^{y^2}=72$ whose solution I know is $x=y=\sqrt 3$, but what are the steps to solve it?
 A: The only solutions are $\pm(\sqrt{3},\sqrt{3})$ and $\pm(\sqrt{1.5},\sqrt{6})$. Proof:
Substitute $a=2x^2, b=y^2$. This becomes $2^a + 2^b = 72$. The relation between $a,b$ becomes $ab=2(xy)^2=18$ and $a,b>0$.
Let $f(x)=2^{x} + 2^{18/x}$. We are interested in positive solutions to $f(x)=72$. Since $f(x)=f(\frac{18}{x})$, we can restrict ourselves to $x \le \sqrt{18}$. 
One solution is $f(3)=2^3 + 2^6=72$, which corresponds to $2x^2=3,y^2=6$, i.e. $\pm(\sqrt{1.5},\sqrt{6})$.
Another solution is recovered - $6=\frac{18}{3}$, which corresponds to $2x^2=6,y^2=3$, i.e. $\pm(\sqrt{3},\sqrt{3})$.
I'll show that $f(x)=72$ has no solutions for $0<x<\sqrt{18}$ other than $x=3$. The proof will be by showing that $f$ is decreasing in that interval:
$$f'(x)=\ln 2 ( 2^x -\frac{18}{x^2} 2^{\frac{18}{x}})$$
For $0<x<\sqrt{18}$,
$$1<\frac{18}{x^2}, 2^{x} < 2^{\frac{18}{x}}$$
So $2^x < \frac{18}{x^2}2^{\frac{18}{x}}$, proving $f' <0$.
A: The first attempt to try to solve a system of two (or more variables) is naturally assumed that a possible solution is one in which all variables are equal. 
Then set $x=y=t$ and solve de equation
\begin{cases}
xy=t\cdot t=3\\
4^{x^2}+2^{y^2}=\big(2^{t^2}\big)^2+2^{t^2}=72
\end{cases}
By $t^2=3$ we have $t=\pm\sqrt{3}$. Substituting into  the another equation we have  that  two solutions are $(x,y)=(+\sqrt{3},+\sqrt{3})$ and  $(x,y)=(-\sqrt{3},-\sqrt{3})$.
Or by $\big(2^{t^2}\big)^2+2^{t^2}=72$ whe have $2^{t^2}=\frac{-1+\sqrt{289}}{2}=8$. 
To further investigate other solutions you might assume
$$
x=t+s, 
\\
y=t-s.
$$
