Solve for "x" and "y"

Question

What would be the easiest way to solve the following set of equations:$$ x + y^2 = 7 $$$$ x^2 + y = 11$$

I've been trying substitution method but end up in a $4$th degree bi-quadratic equation. Thanks for your time!

EDIT: Final equation: $y^4 - 14y^2 + y + 38 = 0$. I'm only concerned about the real solutions. Is it possible to write it as a product of two quadratic equations?

Answer 1

In finding real solutions, consider substituting your first equation into your second, that is, you will have $$x^2\pm\sqrt{7-x}=11,$$ and so $$x^2-11=\pm\sqrt{7-x}.$$ Square both sides: $$x^4-22x^2+121=7-x.$$ Algebraically rearranging, $$x^4-22x^2+x+114=0.$$ Rational root test will tell you that $x=3$ is a root, and hence a solution. WolframAlpha verifies this. Long division or synthetic division reduces our equation to $$(x-3)(x^3+3x^2-13x-38)=0.$$ We have $x=3=0$, which gives us $x=3$, and $$x^3+3x^2-13x-38=0.$$ The latter equation can be solved using Cardano's formula and find the remaining real roots.

Anyway, what we can do for now: substitute the $x=3$ in into either of your original equations to find your $y$, which should be $y=2$.

NB: There are three more real roots of $x$ that solve the fourth-degree polynomial, but those roots are irrational since they are not found in any possible roots of the rational root test.

Answer 2

First note that your two original equations represents two parabolas with orthogonal axes, so they have four intersection points, i.e. the systems has four solutions.

Your equation $y^4-14y^2+y+38=0$ has the obvious solution $y=2$ so it factorize in: $$ (y-2)(y^3+2y^2-10y-19)=0 $$ Since a cubic has always a real solution there are at least two real solution of your quartic.

Now you can study the function $f(y)=y^3+2y^2-10y-19$ . The derivative is $f'=3y^2+4y-10 $ and you can solve the second degree equation for stationary points, finding two real solutions $y=\dfrac{-2\pm\sqrt{34}}{3}$.

With a bit of work you can see that the function has a positive maximum and a negative minimum, so it must have three real solutions as we expected.