I have a problem where I am to find the intersection between the two curves
$$
\begin{cases}
x^2 – y^2 = 3 \\
xy = 2,
\end{cases}
$$
which I can easily see that the two points $\pm(2,1)$ are the real solutions to this problem, but I don't know how to solve for this systematically. I tried two approaches to solve for it as a quadratic equation:
approach 1 (insert the second equation in the first):
$$
4x^2 – 4y^2 + 4xy = 20 \Leftrightarrow (2x + y)^2 – 5y^2 = 20 \Leftrightarrow x=\dfrac{\pm\sqrt{5y^2 + 20} – y}{2}
$$
approach 2 (substitute $y$ for $x$):
$$
x=\frac{2}{y} \Rightarrow x^2 – \frac{4}{x^2} = 3 \Leftrightarrow x^4 – 3x^2 – 4 = 0
$$
But I don't know how to continue from here. I found this math.stackexchange question where they solve a similar equation using these two approaches, but they don't end up with a constant under the root since they have $0$ on the right-hand side. Also, since that equation don't have any real solutions, I don't get how to apply it to this equation.
How do I solve this kind of equation systematically? (not 100% sure what type of the equation it is)
Edits:
Corrected substitution from incorrectly substituting $y$ for $y$. I had arrived at this step earlier, but I still don't know how to solve the equation.
Best Answer
Multiply the first equation by $x^2$ and substitute $xy$ from the second:
$$x^4-x^2y^2-3x^2=x^4-4-3x^2=0.$$
This is a biquadratic equation, which gives the roots $x^2=4$ and $x^2=-1$. If you are only interested in the real solutions,
$$x=\pm2,y=\frac2x.$$