Respuesta :
I: x+y+z=72
II: y=x+1
III: z=x+2
-> substitute y and z in I with the definitions II and III:
x+(x+1)+(x+2)=72
3x+3=72
3x=69
x=23
-> insert x into II and III
y=x+1=23+1=24
z=x+2=23+2=25
-> x+y+z=23+24+25=72
II: y=x+1
III: z=x+2
-> substitute y and z in I with the definitions II and III:
x+(x+1)+(x+2)=72
3x+3=72
3x=69
x=23
-> insert x into II and III
y=x+1=23+1=24
z=x+2=23+2=25
-> x+y+z=23+24+25=72