Question

Determinati nr rationale pozitive x ,y , z care sunt direct prportionale nu nr 5 , 6 , 10 ,, 2x+3y-2Z=128

Answer 1

x/5=k    x=5k
y/6=k    y=6k
z/10=k   z=10k

2*5k+3*6k-2*10k=128
10k+18k-20k=128
8k=128
k=16

x=5*16=80
y=6*16=96
z=10*16=160

Answer 2

x ,y , z care sunt direct prportionale nu nr 5 , 6 , 10 ⇒$\frac{x}{5}$=$\frac{y}{6}$=$\frac{z}{10}$=k⇒
$\frac{x}{5}$=k⇒ x=5·k
$\frac{y}{6}$=k⇒ y=6·k
$\frac{z}{10}$=k⇒ z=10·z
2x+3y-2Z=128

2·5·k+3·6·k-2·10·k=128
10·k+18·k-20·k=128
28·k-20·k=128
8·k=128
k=128:8
k=16

x=5·k⇒x=5·16=80    x=80
y=6·k=6·16=96        y=96
z=10·k=10·16=160    z=160