Răspuns :
Notam: bilele rosii cu a, pe cele albastre cu b, pe cele albe cu c si pe cele galbene cu d;
a+b+c+d=x;
c+d=138; c=d=69 bile;
a=x/6+12=(x+72)/6; b=(x-a)/4+3=(x-x/6-12)/4+3=(5x-72)/24+3; inlocuim in ecuatia de baza: (x+72)/6+(5x-72)/24+3+138=x ⇒ 4x+288+5x-72+141·24=24x ⇒ 9x+216+3384=24x ⇒ 15x=3600, x=240 (nr total de bile).
a=x/6+12=52 bile; b=240-52-138=50 bile.
Raspuns: rosii=52, albastre=50, albe=69, galbene=69.
a+b+c+d=x;
c+d=138; c=d=69 bile;
a=x/6+12=(x+72)/6; b=(x-a)/4+3=(x-x/6-12)/4+3=(5x-72)/24+3; inlocuim in ecuatia de baza: (x+72)/6+(5x-72)/24+3+138=x ⇒ 4x+288+5x-72+141·24=24x ⇒ 9x+216+3384=24x ⇒ 15x=3600, x=240 (nr total de bile).
a=x/6+12=52 bile; b=240-52-138=50 bile.
Raspuns: rosii=52, albastre=50, albe=69, galbene=69.
x = nr. total de bile
bile rosii: x/6+12; rest x-(x/6+12)=5x/6-12
bile albastre: 1/4(5x/6-12)=5x/24, noul rest 5x/6-12-5x/24=5x/8-12
5x/8-12=138⇒x=240 bile in total
bile rosii 52
bile albastre 50
biloe albe = bile galbene = 138/2=69
bile rosii: x/6+12; rest x-(x/6+12)=5x/6-12
bile albastre: 1/4(5x/6-12)=5x/24, noul rest 5x/6-12-5x/24=5x/8-12
5x/8-12=138⇒x=240 bile in total
bile rosii 52
bile albastre 50
biloe albe = bile galbene = 138/2=69