Simple but hard 2 by 2 system in $x$ and $y$

Question

Is there a systematic way of solving this system, analytically?

$$\begin{cases} x \ + \ y^2=11\\ x^2+y\ \ =\ 7\\ \end{cases} $$

I mean, other than brute-force.

Answer 1

You have tagged the question diophantine-equations which implies you are interested in integer solutions.

For those, subtract the equations and factorise to get $(y-x)(x+y-1)=4$. If $x,y$ are integers, then so are $y-x,x+y-1$, so we must have $y-x=1,x+y-1=4$ leading to $x=2,y=3$ or other combinations like $y-x=2,x+y-1=2$ which fail to give solutions. But we have certainly got the only integer solution.

For real solutions there is no choice but the quartic route. You get $(7-x^2)^2+x-11=0$ or $x^4-14x^2+x+38=0$. Suppose you had not already found the $x=2$ solution. Any rational solution $\frac{a}{b}$ must have $b$ a factor of 38 and $a$ a factor of 1 (the coefficient of $x^4)$. So you try $x=2$ and find it is a solution. You now factorise to get: $(x-2)(x^3+2x^2-10x-19)$. You can repeat the process, but trying $x=\pm1,\pm19$ does not give any more solutions. So you have no choice but to use the standard techniques to solve the cubic to get the irrational solutions. It has in fact got three real roots: $3.13131$, $-3.28319$, $-1.84813$.

Answer 2

You don't specify the allowable values of $x,y$. If they are naturals, clearly $y \gt x$ and it is easy to see $x=2, y=3$ is a solution. If they are reals, you can do $y^2=11-x$, plug that into the other and get a quartic in $x$ that the rational root theorem finds the solution (among others). Is that what you mean by brute force?