Question

Determinati perechile de numere intregi (x,y) cu proprietatea ca $\frac{x-1}{3} = \frac{1}{y+1}$ .

Answer 1

$\frac{x-1}{3}= \frac{1}{y+1}=>(x-1)(y+1)=3.$

Cum x si y sunt intregi, distingem urmatoarele cazuri:

$I. \left \{ {{x-1=1} \atop {y+1=3}} \right. \\ II. \left \{ {{x-1=-1} \atop {y+1=-3}} \right. \\ III. \left \{ {{x-1=3} \atop {y+1=1}} \right. \\ IV. \left \{ {{x-1=-3} \atop {y+1=-1}} \right.$

Solutiile sistemelor sunt urmatoarele:
I. (x,y)=(2,2)
II. (x,y)=(0,-4)
III. (x,y)=(4,0)
IV. (x,y)=(-2,-2)

Solutie: (x,y)∈{(-2,-2);(0,-4);(2,2);(4,0)}.