2021 JMPSC Invitationals Problems/Problem 8

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 -

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2021 JMPSC Invitationals Problems/Problem 8

Contents

Problem

Solution

Solution 2

Solution 3

See also