Difference between revisions of "2023 AMC 12A Problems/Problem 10"
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<cmath>y^3=9y^2</cmath> | <cmath>y^3=9y^2</cmath> | ||
<cmath>y=9, x=27 \rightarrow x+y=\boxed{\textbf{(D)} 36}</cmath> | <cmath>y=9, x=27 \rightarrow x+y=\boxed{\textbf{(D)} 36}</cmath> | ||
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+ | ==Video Solution (Under 3 minutes)== | ||
+ | https://youtu.be/G63hRr9bkzI | ||
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+ | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== |
Revision as of 12:25, 19 November 2023
Contents
Problem
Positive real numbers and
satisfy
and
. What is
?
Solution 1
Because , set
,
(
). Put them in
we get
which implies
. Solve the equation to get
or
. Since
and
are positive,
and
.
~plasta
Solution 2
Let's take the second equation and square root both sides. This will obtain . Solving the case where
, we'd find that
. This is known to be false because both
and
have to be positive, and
implies that at least one of the variables is not positive. So we instead solve the case where
. This means that
. Inputting this value into the first equation, we find:
This means that
. Therefore,
~lprado
Solution 3: Quadratic formula
first expand
consider a=1 b=-2y c=-3y^2
or
and
and
we can see both x and y will be postive in
now do same as solution 2
This means that
. Therefore,
Solution 4: Substitution
Since , we can rewrite the second equation as
Let . The second equation becomes
Substituting this into the first equation, we have
so
.
Hence and
-Benedict T (countmath1)
Solution 5: Difference of Squares
We will use the difference of squares in the second equation.
Since x and y are positive, x+y is non-zero. Thus, .
Substituting into the first equation:
Video Solution (Under 3 minutes)
~Education, the Study of Everything
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.