Difference between revisions of "2023 AMC 12A Problems/Problem 10"
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− | == Solution | + | == Solution 3: Quadratic formula== |
first expand | first expand | ||
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<math>x=\frac{2y\pm\sqrt{16y^2}}{2a}</math> | <math>x=\frac{2y\pm\sqrt{16y^2}}{2a}</math> | ||
− | <math>x=\ | + | <math>\frac{2y+4y^{2}}{2}</math> or <math>\frac{2y-4y^{2}}{2}</math> |
+ | |||
+ | <math>x=y+2y</math> and <math>x=y-2y</math> | ||
+ | |||
+ | <math>x=3y</math> and <math>x=-y</math> | ||
+ | |||
+ | we can see both x and y will be postive in <math>x=3y</math> | ||
+ | |||
+ | now do same as solution 2 | ||
+ | <cmath>y^3 = (3y)^2</cmath> | ||
+ | <cmath>y^3 = 9y^2</cmath> | ||
+ | <cmath>y=9</cmath> | ||
+ | This means that <math>x=3y=3(9)=27</math>. Therefore, <math>x+y=9+27=\boxed{36}</math> | ||
+ | |||
+ | == Solution 4: Substitution== | ||
+ | |||
+ | Since <math>a^2 = |a|^2</math>, we can rewrite the second equation as <math>(x-y)^2=4y^2</math> | ||
+ | |||
+ | Let <math>u=x+y</math>. The second equation becomes | ||
+ | |||
+ | |||
+ | <cmath>(u-2y)^2 = 4y^2</cmath> | ||
+ | <cmath>u^2 - 4uy = 0</cmath> | ||
+ | <cmath>u = 4y</cmath> | ||
+ | <cmath>x+y = 4y</cmath> | ||
+ | <cmath>x = 3y.</cmath> | ||
+ | |||
+ | Substituting this into the first equation, we have | ||
+ | |||
+ | <cmath>y^3 = (3y)^2,</cmath> | ||
+ | so <math>x = 9</math>. | ||
+ | |||
+ | Hence <math>x = 27</math> and <math>x + y = \boxed{\textbf{(D)} 36}.</math> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | == Solution 5: Difference of Squares== | ||
+ | |||
+ | We will use the difference of squares in the second equation. | ||
+ | |||
+ | <cmath>(y-x)^2=4y^2</cmath> | ||
+ | <cmath>(y-x)^2-(2y)^2=0</cmath> | ||
+ | <cmath>(y-x-2y)(y-x+2y)=0</cmath> | ||
+ | <cmath>-(x+y)(3y-x)=0</cmath> | ||
+ | |||
+ | Since x and y are positive, x+y is non-zero. Thus, <cmath>3y=x</cmath>. | ||
+ | |||
+ | Substituting into the first equation: | ||
+ | |||
+ | <cmath>y^3=x^2</cmath> | ||
+ | <cmath>y^3=9y^2</cmath> | ||
+ | <cmath>y=9, x=27 \rightarrow x+y=\boxed{\textbf{(D)} 36}</cmath> | ||
+ | |||
+ | ==Video Solution (easy to digest) by Power Solve== | ||
+ | https://youtu.be/YXIH3UbLqK8?si=ia236Rh1UU681RNT&t=1279 | ||
+ | |||
+ | ==Video Solution (⚡ Under 3 minutes ⚡)== | ||
+ | https://youtu.be/G63hRr9bkzI | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/OqlerYo-uPU | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
==See also== | ==See also== |
Latest revision as of 19:35, 10 June 2024
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 (easy to digest) by Power Solve
https://youtu.be/YXIH3UbLqK8?si=ia236Rh1UU681RNT&t=1279
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.