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

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If we move every term dependent on <math>x</math> or <math>y</math> to the LHS, we get <math>x^2 - 10x + y^2 + 6y = -34</math>. Adding <math>34</math> to both sides, we have <math>x^2 - 10x + y^2 + 6y + 34 = 0</math>. Notice this is a circle with radius <math>0</math>, which only contains one point. We can split the <math>34</math> into <math>25</math> and <math>9</math> to get <math>(x - 5)^2 + (y + 3)^2 = 0</math>. So, the only point is <math>(5, -3)</math>, so the sum is <math>5 + (-3) = 2 \implies \boxed{\textbf{(B)}}</math>
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If we move every term dependent on <math>x</math> or <math>y</math> to the LHS, we get <math>x^2 - 10x + y^2 + 6y = -34</math>. Adding <math>34</math> to both sides, we have <math>x^2 - 10x + y^2 + 6y + 34 = 0</math>. Notice this is a circle with radius <math>0</math>, which only contains one point. We can split the <math>34</math> into <math>25</math> and <math>9</math> to get <math>(x - 5)^2 + (y + 3)^2 = 0</math>. So, the only point is <math>(5, -3)</math>, so the sum is <math>5 + (-3) = 2 \implies \boxed{\textbf{(B)}}</math>. ~ asdf334

Revision as of 13:24, 27 June 2019

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 $x + y$?

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


If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$. Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$. Notice this is a circle with radius $0$, which only contains one point. We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$. So, the only point is $(5, -3)$, so the sum is $5 + (-3) = 2 \implies \boxed{\textbf{(B)}}$. ~ asdf334

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