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

Revision as of 11:42, 7 April 2013

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

If we complete the square after bringing the $x$ and $y$ terms to the other side, we get $(x-5)^2 + (y+3)^2 = 0$ . Squares of real numbers are nonnegative, so we need both $(x-5)^2$ and $(y+3)^2$ to be $0$ . This obviously only happens when $x = 5$ and $y = -3$ . $x+y = 5 + (-3) = \boxed{\textbf{(B) }2}$

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2013 AMC 12B Problems|2013 AMC 12B #6]] and [[2013 AMC 10B Problems|2013 AMC 10B #11]]}}
 ==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>?

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

Revision as of 11:42, 7 April 2013

Problem

Solution

See also