Question

gaseste 4 numere  naturale,stiind ca triplul sumei lor este 36;daca primul numar il  inmultim cu  4,al doilea il inmultim cu 2,la al treilea adaugam 1 si din al patrulea scadem 2,atunci obtinem numere egale

Answer 1

a+b+c=12
a*4=b*2=c+1=d-2


le facem pe toate in functie de a

deci 4a=2b
2b=4a
b=4a/2
b=2a

4a=c+1
c+1=4a
c=4a-1

d-2=4a
d=4a+2

inlocuim in suma

a+2a+4a-1+4a+2=12
11a=12-2+1
11a=11
a=1

b=2a=> b=2*1=> b=2

c=4a-1=>c=4*1-1=>c=3

d=4a+2=>d=4*1+2=>d=6

Answer 2

(a+b+c+d)*3=36
4a=2b=c+1=d-2

(a+b+c+d)*3=36 => a+b+c+d= 36/3=12

4a=2b => b=4a/2 = 2a

4a=c+1 => c= 4a-1

4a=d-2 => d= 4a+2

a+2a+4a-1+4a+2 = 12 => 11a= 12-2+1 => 11a = 11 => a=1

b=2a => b = 2*1 = 2

c= 4a-1 => c = 4-1 =3

d=4a+2 => d = 4+2 = 6

verificare: a + b + c + d = 12
1 + 2 + 3 + 6 = 12 (corect)