2013 AMC 10B Problem 11

One equation usually cannot pin down two unknowns, so this one must be special. Look for perfect squares.

1. Move everything to the left side:
$$x^2 - 10x + y^2 + 6y + 34 = 0.$$
2. The part $x^2 - 10x$ is the start of $(x - 5)^2 = x^2 - 10x + 25$. The part $y^2 + 6y$ is the start of $(y + 3)^2 = y^2 + 6y + 9$.
3. The two missing pieces are $25 + 9 = 34$, which is exactly the number we already have. So the equation is
$$(x^2 - 10x + 25) + (y^2 + 6y + 9) = 0, \quad\text{that is,}\quad (x - 5)^2 + (y + 3)^2 = 0.$$
4. The square of a real number is never negative. Two numbers that are each $0$ or more can only add up to $0$ if both are $0$. So $x - 5 = 0$ and $y + 3 = 0$, which gives $x = 5$ and $y = -3$.
5. Check in the original equation: the left side is $25 + 9 = 34$; the right side is $50 - (-18) - 34 = 50 + 18 - 34 = 34$. They match.
6. So $x + y = 5 + (-3) = 2$.

The answer is $\boxed{\textbf{(B)}\ 2}$.

When one equation has two unknowns but the problem expects a single answer, look for a way to write it as a sum of squares equal to zero. Here the constant $34 = 25 + 9$ was the clue: it is exactly what completing both squares needs.

Bring all terms to the left and group by variable:
$$x^2 - 10x + y^2 + 6y + 34 = 0.$$
Complete the square in each variable. Since $x^2 - 10x = (x-5)^2 - 25$ and $y^2 + 6y = (y+3)^2 - 9$, the equation becomes
$$(x-5)^2 + (y+3)^2 - 25 - 9 + 34 = 0 \quad\Longrightarrow\quad (x-5)^2 + (y+3)^2 = 0.$$
A sum of two squares of real numbers is zero only when both squares are zero, so $x = 5$ and $y = -3$. Then $x + y = 2$.

The answer is $\boxed{\textbf{(B)}\ 2}$.