Question

 1 pe x + 1 pe y=1 pe 3 
aflati x si y,vreau rezolvare intreaga

Answer 1

x(y-3)=3y
y=3x/(x-3)
x-3 diferit de zero
Dam valori
x=-6 y=-2
x=2 y=-6
x=4 y=12
x=12 y=4
x=6 y=6

Answer 2

Rezolvam ecuatia pentru x, y∈ ℤ 

Conditii de existenta a ecuatiei :  x ≠ 0;   y ≠ 0

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{3}\Rightarrow \dfrac{1}{y}=\dfrac{1}{3}-\dfrac{1}{x} \Rightarrow \dfrac{1}{y} = \dfrac{x-3}{3x} \Rightarrow y = \dfrac{3x}{x-3}\ \ \ (1)$

[tex]y\in \mathbb{Z} \stackrel{(1)}{\Longrightarrow} \dfrac{3x}{x-3}\in\mathbb{Z} \Rightarrow x-3|3x\ \ \ (2) \\\;\\ Dar,\ \cdot\ (x-3)|(x-3)\Rightarrow x-3|(x-3)\cdot3 \Rightarrow x-3|3x-9\ \ \ (3) \\\;\\ (2), \ (3) \Rightarrow x-3| 9 \Rightarrow x-3 \in D_9 \Rightarrow x-3\in \{-9, -3, -1, 1, 3, 9\}|_{+3} \\\;\\ \Rightarrow x\in\{-6, 0, 2, 4, 6, 12\}.\ Dar\ x\ne0\Rightarrow x\in\{-6, 2, 4, 6, 12\} \ \ (4) \\\;\\ (1), (4) \Rightarrow (x,y)\in\{(-6, 2),\ (2, -6),\ (4, 12),\ (6, 6),\ (12,4)\} .[/tex]