Question

Din suma a trei nr.consecutive,daca scadem diferenta dintre al treilea si primul nr,obtinem 40.
Care sunt nr.?

Answer 1

$(x+y+z)-(z-x)=40 \\ y=x+1 \\ z=y+1=x+2 \\ (x+x+1+x+2)-(x+2-x)=40 \\ (3x+3)-2=40 \\ 3x+1=40 \\ 3x=40-1 \\ 3x=39 \\ x=\frac{39}{3} \\ \boxed{x=13} \\ y=x+1=13+1 \\ \boxed{y=14} \\ z=y+1=14+1 \\ \boxed{z=15}$

Answer 2

x+0,x+1,x+2=
consecutive
(x+0+x+1+x+2)-(x+2-x+0)
=40
3x+3-2=40
3x+1=40
3x+0=39
x+0=13
x+1=14
x+2=15