# 2023 AMC 12A Problems/Problem 10

## 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, $x+y=9+27=\boxed{36}$

first expand

$(y-x)^2 = 4y^2$

$y^2-2xy+x^2 = 4y^2$

$y^2-2yx+x^2 = 4y^2$

$x^2-2xy-3y^2 = 0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

consider a=1 b=-2y c=-3y^2

$x=\frac{2y\pm\sqrt{(-2y)^2-4(1)(-3y^2)}}{2a}$

$x=\frac{2y\pm\sqrt{16y^2}}{2a}$

$\frac{2y+4y^{2}}{2}$ or $\frac{2y-4y^{2}}{2}$

$x=y+2y$ and $x=y-2y$

$x=3y$ and $x=-y$

we can see both x and y will be postive in $x=3y$

now do same as solution 2 $$y^3 = (3y)^2$$ $$y^3 = 9y^2$$ $$y=9$$ This means that $x=3y=3(9)=27$. Therefore, $x+y=9+27=\boxed{36}$

## Solution 4: Substitution

Since $a^2 = |a|^2$, we can rewrite the second equation as $(x-y)^2=4y^2$

Let $u=x+y$. The second equation becomes

$$(u-2y)^2 = 4y^2$$ $$u^2 - 4uy = 0$$ $$u = 4y$$ $$x+y = 4y$$ $$x = 3y.$$

Substituting this into the first equation, we have

$$y^3 = (3y)^2,$$ so $x = 9$.

Hence $x = 27$ and $x + y = \boxed{\textbf{(D)} 36}.$

-Benedict T (countmath1)

## Solution 5: Difference of Squares

We will use the difference of squares in the second equation.

$$(y-x)^2=4y^2$$ $$(y-x)^2-(2y)^2=0$$ $$(y-x-2y)(y-x+2y)=0$$ $$-(x+y)(3y-x)=0$$

Since x and y are positive, x+y is non-zero. Thus, $$3y=x$$.

Substituting into the first equation:

$$y^3=x^2$$ $$y^3=9y^2$$ $$y=9, x=27 \rightarrow x+y=\boxed{\textbf{(D)} 36}$$

## Video Solution (⚡ Under 3 minutes ⚡)

~Education, the Study of Everything

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)