Fie a, b, c cele 3 numere
b = 40 + [tex] \frac{a}{2} [/tex]
c = 3*a
a + b + c = 1012
a + ( 40 + [tex] \frac{a}{2} [/tex] ) + 3*a = 1012
a + 40 + [tex] \frac{a}{2} [/tex] + 3*a = 1012
a + 3*a + [tex] \frac{a}{2} [/tex] = 1012 - 40
4*a + [tex] \frac{a}{2} [/tex] = 972
( aducem la numitor (comun) si apoi inmultim totul cu 2 ca sa scapam de fractie)
2*4*a + a = 972*2
8*a + a = 1944
9*a = 1944
a = [tex] \frac{1944}{9} [/tex]
a = 216
b = 40 + [tex] \frac{a}{2} [/tex] = 40 + 108 = 148
c = 3*a = 3*216 = 648
Proba:a+b+c=216+148+648=1012 Corect.
