Question

Determinati perechile de numere (a,b) pt care are loc egalitatea
(a-1)/2=3/b+1

Answer 1

(a-1)(b+1)=6
avem posibilitatile
1)
a-1=1 ;a=2
b+1=6 ;b=5
2)
a-1=2 ;a=3
b+1=3 ;b=2
3)
a-1=3; a=4
b+1=2; b=1
4)a-1=6; a=7
b+1=1 ;b=0

Answer 2

(a-1)/2=3/(b+1)<=>(a-1)(b+1)=6; 6=6×1=1×6=2×3=3×2=> a-1=1=>a=2;b+1=6=>b=5;(a=2;b=5);a-1=6=>a=7;b+1=1=>b=0;(a=7;b=0); a-1=2=>a=3;b+1=3=>b=2;(a=3;b=2); a-1=3=>a=4;b+1=2=>b=1;(a=4;b=1). Perechile sunt:{(a=2;b=5),(a=7;b=0),(a=3;b=2),(a=4;b=1).}