KVPY 2015

Subtracting $x^{2} + y = 12$ from $y^{2} + x = 12$ gives $(y - x)(y + x - 1) = 0$ , implying either $y = x$ or $y + x = 1$ .

1. Substituting $y = x$ in $x^{2} + y = 12$ , we get $x^2 + x - 12 = 0$ , $x = 3, -4$ . Hence $(3, 3)$ and $(-4, -4)$ are solutions.

2. Substituting $y = 1 - x$ in $x^{2} + y = 12$ , we get $x^2 - x - 11 = 0$ , $x = (1 + 3\sqrt{5}) / 2$ , $(1 - 3\sqrt{5}) / 2$ . Substituting $x = (1 + 3\sqrt{5}) / 2$ in $y = 1 - x$ gives $y = (1 - 3\sqrt{5}) / 2$ . Substituting $x = (1 - 3\sqrt{5}) / 2$ in $y = 1 - x$ gives $y = (1 + 3\sqrt{5}) / 2$ . Hence $((1 + 3\sqrt{5}) / 2, (1 - 3\sqrt{5}) / 2)$ and $((1 - 3\sqrt{5}) / 2, (1 + 3\sqrt{5}) / 2)$ also satisfy the equations.

This gives us a total of $4$ ordered pairs that satisfy the equations.

Drawing the graph clearly shows 4 intersection points The graph

I don't know if this is right but this is what I did.

1. It's easy to see that x+y=1 That's fine by subtracting y^2 to one side and subtracting y to the other. That gives x+y=1

2. I added both equations giving x^2+y^2+x+y=24 and then I substituted x+y=1 to get x^2+y^2=23

3. I set y=1-x an substituted that into this equation x^2+y^2. After simplifying u get x^2-x-11=0 that tieless two solutions for x. Multiply that by two because same thing would happen if u solved for y instead of x yielding 4 real solutions.

Please tell me if this is a mathematically correct method. Thanks!!!