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If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

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

(2) |z + y| = |y| + |z|

From 1)

Either both X and Y are positive or both are negative. Then only 1) can hold true.

If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.

From 2)

Either both Z and Y are positive or both are negative. Then only 2) can hold true.

If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.

Combining 1) and 2)

When both X and Y are positive then Z is positive. i.e (x)(y + z)>0 When both X and Y are negative then Z is negative i.e (x)(y + z) >0

daemnn bro @ rahulgoyal1986! you're oh a roll!! I agree with C

What's the OA?

rahulgoyal1986 wrote:

judokan wrote:

If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

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

(2) |z + y| = |y| + |z|

From 1)

Either both X and Y are positive or both are negative. Then only 1) can hold true.

If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.

From 2)

Either both Z and Y are positive or both are negative. Then only 2) can hold true.

If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.

Combining 1) and 2)

When both X and Y are positive then Z is positive. i.e (x)(y + z)>0 When both X and Y are negative then Z is negative i.e (x)(y + z) >0

daemnn bro @ rahulgoyal1986! you're oh a roll!! I agree with C

What's the OA?

rahulgoyal1986 wrote:

judokan wrote:

If x, y, and z are nonzero numbers, is (x)(y + z) > 0?

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

(2) |z + y| = |y| + |z|

From 1)

Either both X and Y are positive or both are negative. Then only 1) can hold true.

If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.

From 2)

Either both Z and Y are positive or both are negative. Then only 2) can hold true.

If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.

Combining 1) and 2)

When both X and Y are positive then Z is positive. i.e (x)(y + z)>0 When both X and Y are negative then Z is negative i.e (x)(y + z) >0

Re: DS absolute value 2 (n3.7) [#permalink]
06 Aug 2008, 12:05

judokan wrote:

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. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: If xyz ≠ 0, is x(y + z) >= 0? [#permalink]
08 Feb 2014, 01:11

1

This post received KUDOS

Expert's post

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.

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...