2015 AMC 10A Problem 16

The system is symmetric under swapping $x$ and $y$, so subtract and add the equations.

Subtracting the second from the first:
$$y - x = (x - 2)^2 - (y - 2)^2 = (x - y)(x + y - 4).$$
The left side is $-(x - y)$. Because $x \ne y$, we may divide by $x - y$: $-1 = x + y - 4$, so $x + y = 3$.

Adding the two equations:
$$x + y + 8 = (x - 2)^2 + (y - 2)^2 = x^2 + y^2 - 4(x + y) + 8.$$
Cancel the $8$ and substitute $x + y = 3$:
$$x^2 + y^2 = (x + y) + 4(x + y) = 5(x + y) = 15.$$

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

For a symmetric pair of equations, the sum and difference are the natural moves: the difference factors through $x - y$ (which the hypothesis $x \ne y$ lets you cancel) and the sum is symmetric, so it can be written in terms of $x + y$ and $x^2 + y^2$. You never need the individual values of $x$ and $y$, which in fact are the irrational roots of $t^2 - 3t - 1 = 0$.

Having $x + y = 3$, expand the first equation: $y + 4 = x^2 - 4x + 4$, so $x^2 = 4x + y = 3x + (x + y) = 3x + 3$. Similarly $y^2 = 3y + 3$. Adding, $x^2 + y^2 = 3(x + y) + 6 = 15$.

• Dividing by $x - y$ without noticing the sign flip ($y - x = -(x - y)$), which gives $x + y = 5$ and the wrong answer $25$.
• Trying to solve for $x$ and $y$ explicitly; the values are $\frac{3 \pm \sqrt{13}}{2}$ and the arithmetic is needlessly messy.