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For any numbers a and b, a*b=a+b-ab If a*b=0, which of the

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For any numbers a and b, a*b=a+b-ab If a*b=0, which of the [#permalink] New post 04 Aug 2004, 13:27
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For any numbers a and b, a*b=a+b-ab
If a*b=0, which of the following CANNOT be a value of b?
a)2
b)1
c)0
d)-1
e)-3/2





Answer is B, why? I don't get it.
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 [#permalink] New post 04 Aug 2004, 16:15
B.
a = b/(b-1)
b cannot be 1, since that will take a to infinity.
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 [#permalink] New post 04 Aug 2004, 18:11
It is B.

a + b - ab = 0
b = a/(a-1)
a/(a-1)=1 --> 0 = -1 So b can be 1
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 [#permalink] New post 05 Aug 2004, 00:31
I am bit confused here:

ab = a+b -ab, where ab = 0
a = -b

Take the values one by one:
1. b = 2 => a = -2 ; ab = -4 != 0
2. b = 1 => a = -1 ; ab = -1 != 0
3. b = 0 => a = 0 ; ab = 0 OK
4. b =-1 => a = 1 ; ab = -1 != 0
5. b = -3/2 => a = 3/2 ; ab = -9/4 != 0

Does a*b mean ab?????
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 [#permalink] New post 05 Aug 2004, 05:37
al1234 wrote:
Does a*b mean ab?????


No, in a*b, * is a function/symbol
and a*b != ab
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 [#permalink] New post 05 Aug 2004, 07:21
don't over think it. Just plug them all in.

a*b=a+b-ab

If b is, for example, 2, and the answer is 0, then the equation would read,

a+2-2a=0
so solving for a, a=2

Try it with 1, and you get

a+1-a=0
So solving, you cancel out the a, and there's no such thing.

So that's the answer. All the others will produce an a that works with the function.
  [#permalink] 05 Aug 2004, 07:21
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