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Is |x|<1? [#permalink] New post 19 Aug 2004, 08:30
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Hello all

Is |x|<1?
i |x+1|=2|x-1|
ii |x-3| is not equal to zero
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 [#permalink] New post 19 Aug 2004, 09:23
These are my least favorite problems.

the answer is c.

If we solve out the first one, and there are several ways to do it, you see that x could be 1/3 or it could be 3. Both work. So it's insufficient.

How do we do it? That's the sucky part. There are two ways. In the first way, you square both sides and solve for x, but when we do we get 0=3x^2 - 10x + 3, which means we've got to go to the quadratic equation, and we should always try to avoid that. But plugging those numbers into it yields both 1/3 and 3.

The other way is to set up two equations based on the original, with the second one showing the negative side of absolute value. The two equations would be:

a) x+1=2(x-1)
b) x+1=-[2(x-1)]

These are based on the original. If the absolute value of x+1 = x(x-1), then we can say the two equations above. Solving both, we get x=1/3 or x=3.

The second one only tells us that x isn't 3, so that's not sufficient.

Together, it must be that x is 1/3, so that's enough info.
  [#permalink] 19 Aug 2004, 09:23
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