Question

1)Afla numerele care satisfac simultan relatiile:
a×(b+1)=3
(b+1)×(c+2)=15
(c+2)×a=5

Urgeeeeent! Coroana!

Answer 1

a×(b+1)=3   ⇒ ⇒ b+1 =3/a
(b+1)×(c+2)=15
(c+2)×a=5  ⇒ c+2 = 5/a
(b+1)×(c+2)=15
 3/a ×  5/a=15
15/a²=15 ⇒ a²=15/15=1² a=1
(c+ 2) ×1=5
c+2=5 ⇒ c=5-2⇒c=3
1×(b+1) =3  ⇒ b+1 =3 ⇒ b=2
Verific 
1× (2+1) =1×3=3
( 2+1) ×( 3+2) =3×5=15
(3+2)×1=5