Difference between revisions of "2013 AMC 12B Problems/Problem 6"
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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> | + | 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> which only happens when <math>x = 5</math> and <math>y = -3</math>. Therefore, <math>x+y = 5 + (-3) = \boxed{\textbf{(B) }2}.</math> |
== See also == | == See also == |
Revision as of 02:28, 20 January 2014
- 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 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
which only happens when
and
. Therefore,
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.