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Difference between revisions of "2013 AMC 12B Problems/Problem 6"

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==Solution==
 
==Solution==
  
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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>
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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> which only happens when <math>x = 5</math> and <math>y = -3</math>. Therefore, <math>x+y = 5 + (-3) = \boxed{\textbf{(B) }2}.</math>
  
 
== See also ==
 
== See also ==

Revision as of 02:28, 20 January 2014

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 $0,$ which only happens when $x = 5$ and $y = -3$. Therefore, $x+y = 5 + (-3) = \boxed{\textbf{(B) }2}.$

See also

2013 AMC 12B (Problems • Answer Key • Resources)
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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