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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which [#permalink]
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bmwhype2 wrote:
For any numbers x and y, x#y = xy - x - y
if x#y=1 which cannot be the value of y?

-2
-1
0
1
2



Y= (X+1)/(X-1)

Now substitute the choices, For Y=1,

X+1 = X-1

which is not possible, that means Y=1 is not possible.
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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which [#permalink]
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walker wrote:
D

xy-x-y=1 ==> y(x-1)-x=1 ==> y(x-1)=(x+1) ==> y=(x+1)/(x-1) ==>

y=(x-1+2)/(x-1) ==> y=1+2/(x-1)

2/(x-1)<>0 ==> y<>1


similarly x= (y+1)/(y-1)...denominator cannot be 0, hence y cannot be 1.
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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which [#permalink]
D

x#y = xy - x - y where x#y = 1
so, choosing x and y values from option x= 0 and x = 1 then
substitute
1 = 0* 1 - 0 - 1 = -1 , where equation is # if y = 1 ,

seeing with x= 0 and y = -1 ,then yes 1 = 1, same as other too..
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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which [#permalink]
Bunuel wrote:
bmwhype2 wrote:
For any numbers \(x\) and \(y\), \(x@y=xy-x-y\). If \(x@y=1\), which of the following cannot be the value of \(y\) ?

A. -2
B. -1
C. 0
D. 1
E. 2

M23-33


Given \(xy-x-y=1\), which is the same as \((1-x)(1-y)-1=1\) or \((1-x)(1-y)=2\). Now, if \(y=1\) then \((1-x)(1-1)=0\neq{2}\), so in order the given equation to hold true \(y\) cannot equal to 1.

Answer; D.


Hey Bunuel,
What are the guidelines to factor \(xy-x-y=1\) as \((1-x)(1-y)-1\)? Can you please provide a link of study or similar problems? I'm unable to understand.
Thanks,
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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which [#permalink]
prsnt11 wrote:
Bunuel wrote:
bmwhype2 wrote:
For any numbers \(x\) and \(y\), \(x@y=xy-x-y\). If \(x@y=1\), which of the following cannot be the value of \(y\) ?

A. -2
B. -1
C. 0
D. 1
E. 2

M23-33


Given \(xy-x-y=1\), which is the same as \((1-x)(1-y)-1=1\) or \((1-x)(1-y)=2\). Now, if \(y=1\) then \((1-x)(1-1)=0\neq{2}\), so in order the given equation to hold true \(y\) cannot equal to 1.

Answer; D.


Hey Bunuel,
What are the guidelines to factor \(xy-x-y=1\) as \((1-x)(1-y)-1\)? Can you please provide a link of study or similar problems? I'm unable to understand.
Thanks,


I second this question - can someone please provide some guidance? Thanks!!
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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which [#permalink]
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happyface101 wrote:
prsnt11 wrote:

Hey Bunuel,
What are the guidelines to factor \(xy-x-y=1\) as \((1-x)(1-y)-1\)? Can you please provide a link of study or similar problems? I'm unable to understand.
Thanks,


I second this question - can someone please provide some guidance? Thanks!!


There are no guidelines as such to factorization of \(xy-x-y=1\) as \((1-x)(1-y)-1\)
Since the question is asking us to find the values that y cannot take, we are trying to make an equation in terms of y.
If (1-x)(1-y) = 0. This means either x = 1 or y =1

By bringing the equation in the form \((1-x)(1-y)\) = 2
We have proved that (1-x)(1-y) ≠ 0'

Hence y cannot take the value 1

Does this help?
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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which [#permalink]
TeamGMATIFY wrote:
happyface101 wrote:
prsnt11 wrote:

Hey Bunuel,
What are the guidelines to factor \(xy-x-y=1\) as \((1-x)(1-y)-1\)? Can you please provide a link of study or similar problems? I'm unable to understand.
Thanks,


I second this question - can someone please provide some guidance? Thanks!!


There are no guidelines as such to factorization of \(xy-x-y=1\) as \((1-x)(1-y)-1\)
Since the question is asking us to find the values that y cannot take, we are trying to make an equation in terms of y.
If (1-x)(1-y) = 0. This means either x = 1 or y =1

By bringing the equation in the form \((1-x)(1-y)\) = 2
We have proved that (1-x)(1-y) ≠ 0'

Hence y cannot take the value 1

Does this help?



This doesn't help bc I still don't know the thought process of turning xy−x−y=1 into (1−x)(1−y)−1. Everything else after that make sense to me.
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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which [#permalink]
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happyface101 wrote:

This doesn't help bc I still don't know the thought process of turning xy−x−y=1 into (1−x)(1−y)−1. Everything else after that make sense to me.


There are practically four different methods to factorise second-order expressions:

1. Take out common factor
2. Grouping
3. Difference of two squares
4. Middle term breaking

This particular example falls under category 2. If you would like further detailing on these 4 methods, you may either refer to a Maths text book of class 8 or write back to me.
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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which [#permalink]
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bmwhype2 wrote:
For any numbers \(x\) and \(y\), \(x@y=xy-x-y\). If \(x@y=1\), which of the following cannot be the value of \(y\) ?
A. -2
B. -1
C. 0
D. 1
E. 2
M23-33


1 = xy - x - y
take y to LHS and take 1 to RHS
y = xy - x -1
y=x(y-1)-1
Pay attention to the term (y-1) ; if we take the value of y = 1 as given in option D then the term (y-1) will become 0
y=x(y-1)-1 {take y=1}
1=x(1-1)-1
1=x(0)-1
1=-1 THIS DOESNT MAKE SENSE.
SO Y = 1 cannot be the value of y

ANSWER IS D

Originally posted by LogicGuru1 on 26 Jul 2016, 05:28.
Last edited by LogicGuru1 on 26 Jul 2016, 08:26, edited 1 time in total.
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Re: For any numbers x and y, x#y = xy - x - y if x#y=1 which [#permalink]
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bmwhype2 wrote:
For any numbers \(x\) and \(y\), \(x@y=xy-x-y\). If \(x@y=1\), which of the following cannot be the value of \(y\) ?

A. -2
B. -1
C. 0
D. 1
E. 2

M23-33

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