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m04 #24

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Re: m04 #24 [#permalink] New post 13 Oct 2012, 00:35
gmatchase wrote:
Quote:
Is x = 0 ?

1. xy = x
2. x + y = y


For Option1, which of the below method need to be followed?

Method1 ---- xy = x ----> xy-x = 0 ----> x(y-1) = 0 ----> x = 0, y = 1 ---- I thought this method is always true in determining a value
Method2 ---- xy = x - Substitute values x=0 & ycan be any value? or x can be any value and y=1?


both method are fine..

methode 1 is beter...

xy=x..y=1...and x is 0..
xy=x=both can b 1..1*1=1..so it cud be 1 or 0..

statement 2.X+y=y...Take y another side...then X=0..
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Re: m04 #24 [#permalink] New post 13 Oct 2012, 04:35
Hi Sanjoo, As you say we need to use the method1 to get one set of values and use method2 to look any other values are possible?
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Re: m04 #24 [#permalink] New post 14 Dec 2012, 10:48
I still don't understand why (1) is not sufficient. I agree that B by itself is sufficient enough. But with (1), how does it matter what the value of y is, x is equal to zero always, right? Please someone help.
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Re: m04 #24 [#permalink] New post 14 Dec 2012, 10:55
Expert's post
sharmila79 wrote:
I still don't understand why (1) is not sufficient. I agree that B by itself is sufficient enough. But with (1), how does it matter what the value of y is, x is equal to zero always, right? Please someone help.


No that's not correct. Notice that if y=1 then x can be any number. For more check here: m04-71403.html#p803443

Hope it helps.
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Re: m04 #24 [#permalink] New post 15 Dec 2012, 09:25
oops! forgot that fact... Thanks Bunuel!
Re: m04 #24   [#permalink] 15 Dec 2012, 09:25
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