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Re: DS absolute value 2 (n3.7) [#permalink]
06 Aug 2008, 12:05
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) |x + y| = |x| + |y|
(2) |z + y| = |y| + |z|
1) x and y should have same sign don't know about z.. it can be +ve or -ve leads mutliple answers.. insuffciient 2) y and z should have same sign don't know about x.. it can be +ve or -ve leads mutliple answers..
combine. x,y,z must have same signs.
(x)(y + z) > 0 always true..
combined sufficient. C
good question. _________________
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Re: If xyz ≠ 0, is x(y + z) >= 0? [#permalink]
08 Feb 2014, 01:11
This post received KUDOS
If xyz ≠ 0, is x (y + z) >= 0?
xyz ≠ 0 means that neither of unknowns is 0.
(1) |y + z| = |y| + |z| --> either both y and z are positive or both are negative, because if they have opposite signs then |y+z| will be less than |y|+|z| (|-3+1|<|-3|+1). Not sufficient, as no info about x.
(2) |x + y| = |x| + |y| --> the same here: either both x and y are positive or both are negative. Not sufficient, as no info about z.
(1)+(2) Either all three are positive or all three are negative --> but in both cases the product will be positive: x(y+z)=positive*(positive+positive)=positive>0 and x(y+z)=negative*(negative+negative)=negative*negative=positive>0. Sufficient.