Determinati numerele necunoscute in cazurile......
1) x+2 supra 3 =y-1 supra 5=z+1 si x + y + z =142
2) 2x+y supra 11 =3y+z supra 14 =z supra 5 si x+2y+z =15
Ajutor ..
Sunt in clasa a 6 a !
x+2/3 = y-1/5 = z+1/4 = k x+2/3 = k x = 3k-2 y-1/5 = k y = 5k+1 z+1/4 = k z = 4k-1 x+y+z = 142 3k-2 +5k+1 + 4k-1 = 142 12k = 142+2-1+1 12k = 144 k = 144:12 k = 12 x = 3k-2 x = 3*12-2 x = 36-2 x = 34 y = 5k+1 y = 5*12+1 y = 60+1 y = 61 z = 4k-1 z = 4*12-1 z = 48-1 z = 47 V: 34+61+47 = 95+47 = 142
