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Algebra Difficulty 5.7 AIME, harder Find the answer

3. Solve the system of equations:

{x+y−2018=(x−2019)⋅yx+z−2014=(x−2019)⋅zy+z+2=y⋅z \left\{\begin{aligned} x+y-2018 & =(x-2019) \cdot y \\ x+z-2014 & =(x-2019) \cdot z \\ y+z+2 & =y \cdot z \end{aligned}\right.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Answer: (2022, 2, 4), (2018, 0, -2).

Solution: Introduce the substitution x~=x−2019\tilde{x}=x-2019. Then the system becomes {x~+y+1=x~⋅y,x~+z+5=x~⋅z,y+z+2=y⋅z.\left\{\begin{array}{l}\tilde{x}+y+1=\tilde{x} \cdot y, \\ \tilde{x}+z+5=\tilde{x} \cdot z, \\ y+z+2=y \cdot z .\end{array} \quad\right. Transform the system to the form: {(x~−1)(y−1)=2,(x~−1)(z−1)=6,(y−1)(z−1)=3.\left\{\begin{array}{l}(\tilde{x}-1)(y-1)=2, \\ (\tilde{x}-1)(z-1)=6, \\ (y-1)(z-1)=3 .\end{array} \quad\right. Make the substitution a=x~−1,b=y−1,c=z−1a=\tilde{x}-1, b=y-1, c=z-1. Then the system becomes:
{ab=2,ac=6,bc=3.\left\{\begin{array}{l}a b=2, \\ a c=6, \\ b c=3 .\end{array}\right.

abc=±6a b c= \pm 6. Using the last equality, we get that the system has two solutions: {a=2,b=1,c=3.{a=−2,b=−1,c=−3.\left\{\begin{array}{l}a=2, \\ b=1, \\ c=3 .\end{array}\left\{\begin{array}{l}a=-2, \\ b=-1, \\ c=-3 .\end{array}\right.\right. Then {x~−1=2,y−1=1,z−1=3.\left\{\begin{array}{l}\tilde{x}-1=2, \\ y-1=1, \\ z-1=3 .\end{array}\right. or {x~−1=−2,y−1=−1,z−1=−3.\left\{\begin{array}{l}\tilde{x}-1=-2, \\ y-1=-1, \\ z-1=-3 .\end{array}\right. Therefore,

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Remark: 1 point for each correct solution found by trial and error.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.