Difference between revisions of "2013 AMC 12B Problems/Problem 6"
(Created page with "==Problem== 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 ...") |
Turkeybob777 (talk | contribs) (Added Solution) |
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<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8</math> | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8</math> | ||
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| + | ==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> | ||
Revision as of 16:40, 22 February 2013
Problem
Real numbers 
and 
satisfy the equation 
. What is 
?
Solution
If we complete the square after bringing the and terms to the other side, we get 
. Squares of real numbers are nonnegative, so we need both 
and 
to be 
. This obviously only happens when 
and 
. 
