Va rog sa ma ajutati: Impartiti nr. 50 in parti direct proportionale cu 1 cu 3 si cu 6.
Impartiti nr 58 in parti invers proportionale cu 1 cu 3 si cu 5 .

Răspuns :

a) x+y+z=50
{x,y,z} dp {1,3,6}
x=y/3=z/6=k
x=k
y/3=k⇒y=3k
z/6=k⇒z=6k
k+3k+6k=50
10k=50
k=50/10⇒k=5
x=5
y=3·5⇒y=15
z=5·6⇒z=30
b)a+b+c=58
{a,b,c} ip{1,3,5}
a=3b=5c=k
a=k
3b=k⇒b=k/3
5c=k⇒c=k/5
k+k/3+k/5=58
(15k+5k+3k)15=58
23k/15=58
k=58·15/23
k=37,83
a=37,83
b=37,83/3⇒b=12,61
c=37,83/5⇒c=7,56