2013 AMC 12B Problems/Problem 6

Revision as of 22:34, 22 November 2013 by Fadebekun (talk | contribs) (See also)
The following problem is from both the 2013 AMC 12B #6 and 2013 AMC 10B #11, so both problems redirect to this page.

Problem

Real numbers $x$ and $y$ satisfy the equation $x^2 + y^2 = 10x - 6y - 34$. What is $x + y$?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

Solution

If we complete the square after bringing the $x$ and $y$ terms to the other side, we get $(x-5)^2 + (y+3)^2 = 0$. Squares of real numbers are nonnegative, so we need both $(x-5)^2$ and $(y+3)^2$ to be $0$. This obviously only happens when $x = 5$ and $y = -3$. $x+y = 5 + (-3) = \boxed{\textbf{(B) }2}$

See also

2013 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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