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07 Jun 2007, 07:59
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*** jet1445, thanks to add a clear tittle about the content of your question... I have done it for u, Fig 
***
6. Is lxl = y – z ?
(1) x + y = z
(2) x < 0
***
6. Is lxl = y – z ?
(1) x + y = z
(2) x < 0
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07 Jun 2007, 08:29
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(C) for me 
lxl = y – z ?
implies that y - z >= 0 <=> y >= z ?
From 1
x + y = z
<=> x = z - y
=> |x| = |z-y| = |-(y-z)| = |-1|*|y-z| = |y-z|
So,
o If y >= z, then |x| = y - z
o If y < z, then |x| = -(y - z)
INSUFF.
From 2
x < 0 ..... but we have no relationship with x, y and z.
INSUFF.
Combining (1) with (2)
We have:
o If y >= z, then |x| = y - z
o If y < z, then |x| = -(y - z)
and
o x < 0
So, as x = z - y, then z - y < 0 thus, y > z.
We are left with y > z and |x| = y - z
SUFF.
lxl = y – z ?
implies that y - z >= 0 <=> y >= z ?
From 1
x + y = z
<=> x = z - y
=> |x| = |z-y| = |-(y-z)| = |-1|*|y-z| = |y-z|
So,
o If y >= z, then |x| = y - z
o If y < z, then |x| = -(y - z)
INSUFF.
From 2
x < 0 ..... but we have no relationship with x, y and z.
INSUFF.
Combining (1) with (2)
We have:
o If y >= z, then |x| = y - z
o If y < z, then |x| = -(y - z)
and
o x < 0
So, as x = z - y, then z - y < 0 thus, y > z.
We are left with y > z and |x| = y - z
SUFF.
07 Jun 2007, 10:46
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(C)
lxl = y – z ?
implies that x = (y - z) FOR ALL x > 0
and -x = (y - z) i.e. x = (z - y) for all x < 0.
From 1
x + y = z
<x> 0 or x < 0.
INSUFF.
From 2
x < 0 ..... but we have no relationship with x, y and z.
INSUFF.
Combining (1) with (2)
We have:
x = (z - y) for all x < 0
i.e. lxl = y – z.
SUFF.
lxl = y – z ?
implies that x = (y - z) FOR ALL x > 0
and -x = (y - z) i.e. x = (z - y) for all x < 0.
From 1
x + y = z
<x> 0 or x < 0.
INSUFF.
From 2
x < 0 ..... but we have no relationship with x, y and z.
INSUFF.
Combining (1) with (2)
We have:
x = (z - y) for all x < 0
i.e. lxl = y – z.
SUFF.
