Help me for my work please

Let David has $x, Carl has $(x+9), and Hilbert has (x+9)/3.

x+(x+9)+(x+9)/3=61
2x+(x+9)/3=52
6x+x+9=156
7x=147
x=21

Thus, David has $21

Carl has $(21+9), ie, $30

Hilbert has (21+9)/3, ie. $10

2013-10-28 20:13:32 補充：
其實設2個未知數、3個未知數都得；
不過設得愈少未知數，就愈簡單，也計得愈快。

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