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Fei Fei Math Q11

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Fei Fei Math Q11

文章JS » 2005-05-04 17:40

DS.Q11
Is point (r,s) on line y=2x+3 ?
a, (2r-s+3)(4r+2s-6)=0
b, (3r+2s-5)(2r-s+3)=0

Ans. E

But I think it should be C
because from a we got two lines
2r-s+3=0 or 4r+2s-6=0
and from b we got
3r+2s-5=0 or 2r-s+3=0
and the 2r-s+3=0 is the only line on both statements
so the answer should be C

or I miss something here?

also Q14 is kinda strange??..... Does Fei Fei math worth reading? 8-)
JS
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文章: 81
註冊時間: 2005-04-07 16:39
來自: London

文章ndxica » 2005-05-05 15:27

no.....只要r,s 同時符合 (4r+2s-6)=0 and (3r+2s-5) 就沒必要 (2r-s+3)=0
了 我是懶得算啦 不過最好還是把上面的r,s算出來 確定他不會lie on y=2x+3
總之 這題真是細心加細心 算是難題
大家一起家加油
ndxica
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文章: 15
註冊時間: 2005-05-05 15:06
來自: Taipei

文章JS » 2005-05-05 15:57

Hi ndxica I follow your suggestion and sort out the volue of (r,s).
but now I got the answer D

From a we know that
2r-s+3=0
4r+2s-6=0
so, s=3, r=0
put(0,3) in the line y=2x+3 we get 3=0+3
so a is sufficient

now b
3r+2s-5=0
2r-s+3=0
r=-1/7, s=19/7
put(-1/7,19/7) into y=2x+3 we get 19/7=-2/7+3
so b is sufficient too

now I am confused!!
JS
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文章: 81
註冊時間: 2005-04-07 16:39
來自: London

文章Jung » 2005-05-05 21:41

From A, it's either 2r-s+3=0 or 4r+2s-6=0 , can not combine both equations together
You can only have the variables instead of the exactly number

The same appears in B

Combining A and B, just like Ndxica mentioned

So the answer is E.
頭像
Jung
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文章: 72
註冊時間: 2004-10-29 10:30


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