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Source: — Data Sufficiency |
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by Night reader » Mon Jan 10, 2011 4:25 pm
GHong14 wrote:Image
xy+z=x(y+2) <=> xy+z=xy+xz <=> z=xz <=> possible solutions x=1 and z=any number OR z=0 and x=any number.

Answer E.
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by Anurag@Gurome » Mon Jan 10, 2011 10:12 pm
.... (xy + z) = x(y + z)
=> (xy + z) = (xy + xz)
=> z = xz
=> (xz - z) = 0
=> z(x - 1) = 0

Therefore we have two sets of possible solution:
  • 1. z = 0, x = any numbers
    2. x = 1, z = any numbers
Only option E matches with the 2nd solution.

The correct answer is E.
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