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

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==Solution 1==
 
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
 
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
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==Solution 2==
 
If we move every term including <math>x</math> or <math>y</math> to the LHS, we get <cmath>x^2 - 10x + y^2 + 6y = -34.</cmath>  We can complete the square to find that this equation becomes <cmath>(x - 5)^2 + (y + 3)^2 = 0.</cmath>  Since the square of any real number is nonnegative, we know that the sum is greater than or equal to <math>0</math>.  Equality holds when the value inside the parhentheses is equal to <math>0</math>.  We find that <cmath>(x,y) = (5,-3)</cmath> and the sum we are looking for is <cmath>5+(-3)=2 \implies \boxed{\textbf{(B)}}.</cmath>  
 
If we move every term including <math>x</math> or <math>y</math> to the LHS, we get <cmath>x^2 - 10x + y^2 + 6y = -34.</cmath>  We can complete the square to find that this equation becomes <cmath>(x - 5)^2 + (y + 3)^2 = 0.</cmath>  Since the square of any real number is nonnegative, we know that the sum is greater than or equal to <math>0</math>.  Equality holds when the value inside the parhentheses is equal to <math>0</math>.  We find that <cmath>(x,y) = (5,-3)</cmath> and the sum we are looking for is <cmath>5+(-3)=2 \implies \boxed{\textbf{(B)}}.</cmath>  
  
 
{{AMC12 box|year=2013|ab=B|num-b=5|num-a=7}}
 
{{AMC12 box|year=2013|ab=B|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:44, 15 August 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 $x + y$?

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


Solution 1

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

Solution 2

If we move every term including $x$ or $y$ to the LHS, we get \[x^2 - 10x + y^2 + 6y = -34.\] We can complete the square to find that this equation becomes \[(x - 5)^2 + (y + 3)^2 = 0.\] Since the square of any real number is nonnegative, we know that the sum is greater than or equal to $0$. Equality holds when the value inside the parhentheses is equal to $0$. We find that \[(x,y) = (5,-3)\] and the sum we are looking for is \[5+(-3)=2 \implies \boxed{\textbf{(B)}}.\]

2013 AMC 12B (Problems • Answer Key • Resources)
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Problem 5 		Followed by
Problem 7
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All AMC 12 Problems and Solutions

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