2015 AMC 10A Problem 16

The two equations look the same with $x$ and $y$ swapped. For a matched pair like that, subtracting and adding them are the natural moves.

Step 1: subtract to find $x + y$.

1. First equation minus second equation:
$$(y + 4) - (x + 4) = (x - 2)^2 - (y - 2)^2.$$
2. The left side is $y - x$. The right side is a difference of two squares, $A^2 - B^2 = (A - B)(A + B)$ with $A = x - 2$ and $B = y - 2$:
$$(x - 2)^2 - (y - 2)^2 = (x - y)(x + y - 4).$$
3. Write the left side as $y - x = -(x - y)$. So
$$-(x - y) = (x - y)(x + y - 4).$$
4. The problem says $x \ne y$, so $x - y$ is not zero and we can divide both sides by it:
$$-1 = x + y - 4, \quad\text{so}\quad x + y = 3.$$

Step 2: add to find $x^2 + y^2$.

5. Add the two equations:
$$x + y + 8 = (x - 2)^2 + (y - 2)^2 = x^2 - 4x + 4 + y^2 - 4y + 4.$$
6. Cancel the $8$ on both sides and collect terms:
$$x^2 + y^2 = x + y + 4x + 4y = 5(x + y).$$
7. Use $x + y = 3$: $x^2 + y^2 = 5 \times 3 = 15$.

The answer is $\boxed{\textbf{(B)}\ 15}$.

When two equations are mirror images (swap $x$ and $y$ and one turns into the other), subtracting them always produces a factor $x - y$, and the condition $x \ne y$ is exactly what lets you cancel it. Adding them keeps everything in terms of $x + y$ and $x^2 + y^2$. You never need $x$ and $y$ themselves, which is lucky, because they are messy square-root numbers.

Once you know $x + y = 3$, go back to the first equation: $y + 4 = x^2 - 4x + 4$, so $x^2 = 4x + y = 3x + (x + y) = 3x + 3$. In the same way $y^2 = 3y + 3$. Adding, $x^2 + y^2 = 3(x + y) + 6 = 9 + 6 = 15$.