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Re: Is xy > 0 ? (1) |x| + y = |x + y| (2) x + |y| = |x + y| [#permalink]
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chetan2u wrote:
Is \(xy>0\)?

(1) \(|x|+y=|x+y|\)
(2) \(x+|y|=|x+y|\)



self made


Well number plugging method has already been explained so I'll go the algebraic way ;)

For \(xy>0\), both \(x\) & \(y\) have to be of the same sign and \(x\) & \(y\) should not both be equal to \(0\)

Statement 1: RHS of the equation is a mod function hence it will always be \(≥0\)

so we have \(|x|+y≥0 => |x|≥-y\). For this to be true \(y≤0\) and \(x≤0\) or \(x≥0\). Hence multiple scenarios possible. Insufficient

Statement 2: again the same logic applies \(x+|y|≥0\), so \(|y|≥-x\). For this to be true \(x≤0\) and \(y≤0\) or \(y≥0\). Hence multiple scenarios possible. Insufficient

Combining statement 1 & 2 we know that either \(x≤0\) or \(x≥0\) and \(y≤0\) or \(y≥0\). So both \(x\) & \(y\) can be positive / negative or both can be \(x=y=0\). Multiple values possible. Hence Insufficient

Option \(E\)
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Re: Is xy > 0 ? (1) |x| + y = |x + y| (2) x + |y| = |x + y| [#permalink]
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