# 2021 JMPSC Invitationals Problems/Problem 8

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## Problem

Let $x$ and $y$ be real numbers that satisfy $$(x+y)^2(20x+21y) = 12$$ $$(x+y)(20x+21y)^2 = 18.$$ Find $21x+20y$.

## Solution

We let $a=(x+y)$ and $b=(20x+21y)$ to get the new system of equations $$a^2b=12 \qquad (1)$$ $$ab^2=18 \qquad(2).$$ Multiplying these two, we have $(ab)^3=12 \cdot 18$ or $$ab=6 \qquad (3).$$ We divide $(3)$ by $(1)$ to get $a=2$ and divide $(2)$ by $(1)$ to get $b=3$. Recall that $a=x+y=2$ and $b=20x+21y=3$. Solving the system of equations $$x+y=2$$ $$20x+21y=3,$$ we get $y=-37$ and $x=39$. This means that $$21x+20y=20x+21y+x-y=3+39-(-37)=\boxed{79}.$$ ~samrocksnature

## Solution 2

Each number shares are factor of $6$, which means $(x+y)(20x+21y)=6$, or $x+y=2$ and $20x+21y=3$. We see $y=-37$ and $x=39$, so $39(21)-20(37)=\boxed{79}$

~Geometry285

## Solution 3

Multiplying the equations together, we get $$(x+y)^3(20x+21y)^3=2^3 \cdot 3^3 \implies (x+y)(20x+21y)=6$$Therefore, $$x+y=2 \implies 20x+20y=40$$$$20x+21y=3$$Subtracting the equations, we get $y=-37$ and $x=39$, therefore, $21 (39) - 20 (37) =\boxed{79}$

- kante314 -