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Difference between revisions of "2023 AMC 12A Problems/Problem 10"

(→‎Solution 1)
(→‎Solution 2)
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<cmath>y^3 = (3y)^2</cmath>
 
<cmath>y^3 = (3y)^2</cmath>
 
<cmath>y^3 = 9y^2</cmath>
 
<cmath>y^3 = 9y^2</cmath>
−
<cmath>y=9</cmath>.
+
<cmath>y=9</cmath>
 
This means that <math>x=3y=3(9)=27</math>. Therefore, <math>x+y=9+27=\boxed{36}</math>
 
This means that <math>x=3y=3(9)=27</math>. Therefore, <math>x+y=9+27=\boxed{36}</math>
  

Revision as of 20:51, 9 November 2023

Problem

Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$. What is $x+y$? $\textbf{(A) }12\qquad\textbf{(B) }18\qquad\textbf{(C) }24\qquad\textbf{(D) }36\qquad\textbf{(E) }42$

Solution 1

Because $y^3=x^2$, set $x=a^3$, $y=a^2$ ($a\neq 0$). Put them in $(y-x)^2=4y^2$ we get $(a^2(a-1))^2=4a^4$ which implies $a^2-2a+1=4$. Solve the equation to get $a=3$ or $-1$. Since $x$ and $y$ are positive, $a=3$ and $x+y=3^3+3^2=\boxed{\textbf{(D)} 36}$.

~plasta

Solution 2

Let's take the second equation and square root both sides. This will obtain $y-x = \pm2y$. Solving the case where $y-x=+2y$, we'd find that $x=-y$. This is known to be false because both $x$ and $y$ have to be positive, and $x=-y$ implies that at least one of the variables is not positive. So we instead solve the case where $y-x=-2y$. This means that $x=3y$. Inputting this value into the first equation, we find: \[y^3 = (3y)^2\] \[y^3 = 9y^2\] \[y=9\] This means that $x=3y=3(9)=27$. Therefore, $x+y=9+27=\boxed{36}$

~lprado

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

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