Algebra problem #1258

By the trivial inequality , since $(x+y-40)^2 \geq 0$ and $(x-y-18)^2 \geq 0$ , thus we must have $x + y - 40 = 0$ and $x - y - 18 = 0$ . Adding these two equations we have $2x = 40 + 18 = 58$ . Hence $x = 29$ is the only possible answer.

We check that this gives $y = 11$ , and that $x+ y - 40 = 0, x - y - 18 = 0$ .

I did it in two methods, first is similar to Arron Kau's method which uses trivial inequality. (Check out his great solution)

The second method I used isn't pretty at all, but I would love to share it with fellow Brilliantians. It's quite lengthy and messy.

Here it goes: Expand the given equation, simplify, form a quadratic equation for x. Since x,y are real numbers, the discriminant must be greater than or equal to 0. Solving that inequality yields some value for y. Subbing it back into the quadratic equation we obtained earlier attains the result x=29.

I know its a crappy method but doing it in many ways is (usually?) better. Thanks