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Difference between revisions of "2013 AMC 10B Problems/Problem 11"

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

Revision as of 15:39, 22 February 2013

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 $y = -3$. $x+y = 5 + (-3) = \boxed{\textbf{(B) }2}$

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