Question

determinati perechile de numere intregi (x,y) cu proprietatea ca x-1supra 3=1supray+1

Answer 1

$\frac{x-1}{3} = \frac{1}{y+1} \\ y \neq -1 \\ (x-1)(y+1)=3 \\ (x,y)=(2,2);(4,0);(-2,-2);(0,-4)$

Answer 2

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

$\Rightarrow (x-1;y+1)\in\{(1;3);\ (3;1);\ (-1;-3);\ (-3;-1)\}\Rightarrow$

$(x;y)\in\{(2;2);\ (4;0);\ (0;-4);\ (-2;-2)\}$

Niciuna din valorile gasite pentru y nu anuleaza numitorul. (y≠1)