Difference between revisions of "2001 AMC 10 Problems/Problem 10"
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Since we know every variable's value, we can substitute it in for <math> x+y+z = 4+6+12 = \boxed{\textbf{(D) }22} </math>. | Since we know every variable's value, we can substitute it in for <math> x+y+z = 4+6+12 = \boxed{\textbf{(D) }22} </math>. | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2001|num-b=9|num-a=11}} |
Revision as of 16:20, 16 March 2011
Problem
If ,
, and
are positive with
,
, and
, then
is
Solution
Look at the first two equations in the problem.
and
.
We can say that .
Given , we can substitute
for
and find
.
We can replace y into the first equation.
.
Since we know every variable's value, we can substitute it in for .
See Also
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 |