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# xyz <> 0, is x (y + z) = 0? (1) y + z = y + z (2) x +

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SVP
Joined: 05 Jul 2006
Posts: 1747
Followers: 6

Kudos [?]: 357 [0], given: 49

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22 Oct 2006, 16:48
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xyz <> 0, is x (y + z) = 0?

(1) Â¦ y + z Â¦ = Â¦ y Â¦ + Â¦ z Â¦

(2) Â¦ x + y Â¦ = Â¦ x Â¦ + Â¦ y Â¦
Senior Manager
Joined: 08 Jun 2006
Posts: 337
Location: Washington DC
Followers: 1

Kudos [?]: 54 [0], given: 0

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22 Oct 2006, 18:50
A

1 - Sufficient

x <> 0, y <> 0, z <> 0
Â¦ y + z Â¦ = Â¦ y Â¦ + Â¦ z Â¦ possible only when y and z have same sign â€“ either both positive or both negative.
The expression x (y + z) can be zero only when y + z is 0 because x can never be 0 (As xyz <> 0)
But (y + z) is not 0 because both y and z has same sign.

So x (y + z) <> 0

2 - Insufficient
Apply same logic
Manager
Joined: 31 Aug 2006
Posts: 211
Followers: 1

Kudos [?]: 2 [0], given: 0

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23 Oct 2006, 12:07
If xyz<>0 then none of x,y and z is0.
For x(y+z) =0 , the following must be true
y=-z

Now (2) gives no relation between y and z
Hence answer has to be (A),(C) or (E)
Again from Stat(1) we get y=z, hence y!=-z
Thus (A)
23 Oct 2006, 12:07
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# xyz <> 0, is x (y + z) = 0? (1) y + z = y + z (2) x +

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