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$x^2 + xy =3$ 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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sure. Take $x=y^2 + 2$ and plug into the other one. You get a polynomial equation in $y,$ graph that and see that there are no roots. – Will Jagy Jan 13 '14 at 2:15

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.

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Thanks! I'm not used to these types of problems and you showed your thinking super well. – jd123 Jan 13 '14 at 2:25

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.

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