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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]
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For y=2 the equation gives 1=0 which is impossible.

Answer (B)
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]
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A. plug in 3 for y to get 2x-3x-3=0 or -x=3. This could work
B. plug in 2 for y to get 2x-2-2x=0 or 0=2 and this is not true

Answer B!!
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]
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Answer should be B)

We can reduce 2x - y - xy = 0 to y = 2x/1+x

All numbers apart from y = 2 can be equated correctly.
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]
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Given \(x \diamond y\) = 2x - y - xy.
And \(x \diamond y\) = 0

Let's Put this value in first equation. So, 2x-xy=y => x= y/2-y
Now we can check the given options one by one from the above equation.

a) y=3 => x = -3 Possible
b) y=2 => x= infinite, Y can not be 2.

hence Answer is B --> 2
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]
Means that

2x-y-xy=0
x(2-y)-y=0
x(2-y)=y
so, test every option

A) x(2-3)=3 is possible when x=-3
B) x(2-2)=2 is not possible in any x
C) x(2-0)=0 is possible when x=0
D) x(2--4/3)=-4/3 is possible when x=-4/3:10/3
E) x(2--2)=-2 is possible when x=-1/2

B
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]
Solve the algebric equation 2x-y-xy=0
2x=y+xy
If y is 2 then the equation could not be equal in any case
as 2x=2+2x so (B)
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Re: For any numbers x and y, x<>y = 2x - y - xy. If x<>y = 0, then which [#permalink]
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