$x^2 + xy =3$ and $x - y^2 = 2$? I graphed it and there are no intersections so obviously there is no real number solutions, but is there a "mathier" (read algebraic) way to prove this?
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From the second equation, $x = 2 + y^2$. Substituting this into the first equation gives
$$3 = x^2 + xy = (2 + y^2)^2 + (2 + y^2) y = (2 + y^2)(2 + y^2 + y) $$
Now $2 + y^2 \geq 2$ and $$y^2 + y + 2 = \left(y + \frac 1 2\right)^2 +\frac7 4 \geq \frac 74$$
So the right hand side of the above equation is at least
$$2 \cdot \frac 7 4 = 3.5 \gt 3$$ So there's no solution.
Usually, to show that a solution doesn't exist, you want to come up with a function that looks some thing like $(x+n)^2+m=0$ where $(m\gt0)$. Because $(x+n)^2$ is always greater than $0$, we can say $x$ doesn't exist.