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

Latest revision as of 12:10, 7 April 2013

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-==Problem==
+#REDIRECT [[2013 AMC 12B Problems/Problem 6]]
-Real numbers <math>x</math> and <math>y</math> satisfy the equation <math>x^2 + y^2 = 10x - 6y - 34</math>.  What is <math>x+y</math>?
-<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2  \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 </math>
-==Solution==
-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>
-== See also ==
-{{AMC10 box|year=2013|ab=B|num-b=10|num-a=12}}