TITLE: Solve $xy=3$ and $4^{x^2}+2^{y^2}=72$
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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
$$
x=t+s, 
\\
y=t-s.
$$