Difference between revisions of "2013 AMC 10B Problems/Problem 11"
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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> | 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> | ||
| + | == See also == | ||
| + | {{AMC10 box|year=2013|ab=B|num-b=10|num-a=12}} | ||
Revision as of 17:01, 27 March 2013
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 
. This obviously only happens when 
and 
. 
See also
| 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 | ||
