Question

determinati numerele x y z stiind ca invers proportionale cu 2 4 5 si a+b+c=38

Answer 1

2x=4y=2z = k
2x = k, x = k/2
4y=k, y = k/4
5z = k, z = k/5
a+b+c = 38
k/2+k/4+k/5 = 38
10k+5k+4k = 760
19k = 760
k = 760:19
k = 40
x = k/2 = 40/2 = 20
y = k/4 = 40/4 = 10
z = k/5 = 40/5 = 8

Answer 2

{a,b,c} ip {2,4,5}⇒a/1/2=b/1/4=c/1/5=k⇒a=k/2,b=k/4,c=k/5
a+b+c=38
k/2+k/4+k/5=38 |*20
10k+5k+4k=760
19k=760
k=40
a=k/2=20
b=k/4=10
c=k/5=8