ctm_0011

It is given that (a+b-c)/15=(b+c-a)/9=(a+c-b)/5.
(a) Find a:b:c.
(b) If (a^2+bc)/x=(b^2+ac)/y=(c^2-ab)/z ≠0 where x, y,and z ≠ 0, find the value of (x-y)/(y+z).

thx

firoazen

equal ration thm

你學生

Let (a+b-c)/15=(b+c-a)/9=(a+c-b)/5 = t
...
a=10t
b=12t
c=7t
...

桃子

(a+b-c)/15=(b+c-a)/9=(a+c-b)/5
Multiply the equation by the LCM 45.
3(a+b-c)=5(b+c-a)=9(a+c-b)

Now, you have a system of 3 equations of 3 unknowns.
3(a+b-c)=5(b+c-a)
3(a+b-c)=9(a+c-b)
5(b+c-a)=9(a+c-b)

Then, you can find a:b:c.

[ 本帖最後由 桃子 於 2011-7-11 07:32 PM 編輯 ]

cheungpuiho04

b) You may assume a=1 b=...
then substitute the values into the equation

桃子

引用:

原帖由 cheungpuiho04 於 2011-7-11 08:47 PM 發表
b) You may assume a=1 b=...
then substitute the values into the equation

It will be easier to put a=10, b=12 and c=7 assuming the answer in #3 is right.