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

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==Problem==
 
==Problem==
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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>?
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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>?
  
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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>
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<math>\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8</math>
  
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==Solution==
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==Solution 1==
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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>. 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>. Notice this is a circle with radius <math>0</math>, which only contains one point. 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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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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==Solution 2==
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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> - Honestly
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== Video Solution ==
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https://youtu.be/ba6w1OhXqOQ?t=1810
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~ pi_is_3.14
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==Video Solution==
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https://youtu.be/a-3CAo4CoWc
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–no one
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==See Also==
  
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== See also ==
 
 
{{AMC12 box|year=2013|ab=B|num-b=5|num-a=7}}
 
{{AMC12 box|year=2013|ab=B|num-b=5|num-a=7}}
 
{{AMC10 box|year=2013|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2013|ab=B|num-b=10|num-a=12}}
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[[Category:Introductory Algebra Problems]]
 
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[[Category:Algebraic Manipulations Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:04, 14 July 2021

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 $x+y$?

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

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 $x^2 - 10x + y^2 + 6y + 34 = 0$. We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$. Notice this is a circle with radius $0$, which only contains one point. 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 \[5+(-3)=2 \implies \boxed{\textbf{(B)}}.\] - Honestly

Video Solution

https://youtu.be/ba6w1OhXqOQ?t=1810

~ pi_is_3.14

Video Solution

https://youtu.be/a-3CAo4CoWc

–no one

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
2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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