2a/3 = 3b/5 = 5c/4 = k
2a/3=k, a=3k/2
3b/5=k b=5k/3
5c/4=k. c=4k/5
Inlocuim in a + b + c = 119⇒ 3k/2+5k/3+4k/5=119 (numitor comun 30)
Inmultim relatia cu 30 : 45k+50k+24k=119×30⇒119k=119×30
⇒k=30
a=3k/2=(3×30)/2=90/2=45
b=5k/3=(5×30)/3=50
c=4k/5 =(4×30):5=24
a = 45, b = 50 si c = 24
verificare
a+b+c=119
45+50+24=119
119=119
2a/3 = 3b/5 = 5c/4 = k
a = 3k/2
b = 5k/3
c = 4k/5
3k/2 + 5k/3 + 4k/5 = 119
45k + 50k + 24k = 119k = 3570
=> k = 3570 : 119 = 30
a = 3k/2 = (3 × 30) : 2 = 90/2 = 45
b = 5k/3 = (5 × 30) : 3 = 150 : 3 = 50
c = 4k/5 = (4 × 30) : 5 = 120 : 5 = 24
