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I'm not sure about C (haven't gotten there in my analysis yet, but I know it's not A, here is what i think:

Without the absolute value, y + z = 0 would have to mean that |y| = |z|. Say y = 3 and z = -3. Then |3-3|= |0|, but |3| + |-3| = 6. One way #1 can be true is if y or z = 0. The only way x(y+z) = 0 is if x or (y+z) = 0. We know x ≠ 0 because xyz =0 would be true but the stems says otherwise. The only way #1 can be true and we can answer that x(y+z) = 0 [implied is "all the time"], both y and z must = 0

Because we can find multiple situations that allow |y + z| = |y| + |z| and some of those situations make (y+z) = 0 and others do not, A is insufficient.

sset009 wrote:

If xyz ≠ 0, is x (y + z) = 0?

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

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

I'm not sure about C (haven't gotten there in my analysis yet, but I know it's not A, here is what i think:

Without the absolute value, y + z = 0 would have to mean that |y| = |z|. Say y = 3 and z = -3. Then |3-3|= |0|, but |3| + |-3| = 6. One way #1 can be true is if y or z = 0. The only way x(y+z) = 0 is if x or (y+z) = 0. We know x ≠ 0 because xyz =0 would be true but the stems says otherwise. The only way #1 can be true and we can answer that x(y+z) = 0 [implied is "all the time"], both y and z must = 0

Because we can find multiple situations that allow |y + z| = |y| + |z| and some of those situations make (y+z) = 0 and others do not, A is insufficient.

sset009 wrote:

If xyz ≠ 0, is x (y + z) = 0?

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

Actually that is incorrect. The answer should be A.

xyz != 0. This means x!=0, y!=0 AND z!=0 (!= ---> not equal)

#1 |y+z| = |y| + |z| This is true only if y = z OR y = 0 OR z = 0. We know the latter two are not true. If y = -z, then there is an inequality. Try it with numbers.

So we know from #1 that y!=-z, therefore y+z !=0. So A says no. So answer has to be A or D

#2 |x+y| = |x| + |y| use the same logic above to find that x = y. This doesn't tell us anything about the original question and is irrelevant.

I'm not sure about C (haven't gotten there in my analysis yet, but I know it's not A, here is what i think:

Without the absolute value, y + z = 0 would have to mean that |y| = |z|. Say y = 3 and z = -3. Then |3-3|= |0|, but |3| + |-3| = 6. One way #1 can be true is if y or z = 0. The only way x(y+z) = 0 is if x or (y+z) = 0. We know x ≠ 0 because xyz =0 would be true but the stems says otherwise. The only way #1 can be true and we can answer that x(y+z) = 0 [implied is "all the time"], both y and z must = 0

Because we can find multiple situations that allow |y + z| = |y| + |z| and some of those situations make (y+z) = 0 and others do not, A is insufficient.

sset009 wrote:

If xyz ≠ 0, is x (y + z) = 0?

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

Actually that is incorrect. The answer should be A.

xyz != 0. This means x!=0, y!=0 AND z!=0 (!= ---> not equal)

#1 |y+z| = |y| + |z| This is true only if y = z OR y = 0 OR z = 0. We know the latter two are not true. If y = -z, then there is an inequality. Try it with numbers.

So we know from #1 that y!=-z, therefore y+z !=0. So A says no. So answer has to be A or D

#2 |x+y| = |x| + |y| use the same logic above to find that x = y. This doesn't tell us anything about the original question and is irrelevant.

A.

Hold on

|y + z| = |y| + |z|

If I chose y=7 and z=8 then above is true so statement "This is true only if y = z OR y = 0 OR z = 0" is wrong

Coming to Q answer should be E)

If I chose x=6, y=7 and z=8 then none of conditions satisfy even if you join both the statements

The question asks if x(y+z) = 0? As x cannot be 0 We have to see whether y+z = or ≠ 0

(1) if y and z are of different signs (+ and -)

I y + z I ≠ |y| + |z|

So (1) only tells us that y and z are of same sign

Nothing can proof y + z = 0

So (1) alone is Not sufficient

Look at the two blue bold parts in your message above.

y and z are of same sign, y≠0, z≠0

It is enough to prove that y+z≠0

If y+z=0, then y=-z : either y=z=0 (this is not the case here) or y and z are of different sign (that is not the case here) ==> y+z cannot be equal to 0

Because we can find multiple situations that allow |y + z| = |y| + |z| and some of those situations make (y+z) = 0 and others do not, A is insufficient.

the only way that the above statement is true if both are negative, or both are positive. e.g y = -3, z = -2 |-5| = |-3| + |-2|

or if y = 3, z = 2 |5| = |3| + |2|

addition of two negative or positive numbers will take it furhter away from zero. if the are different signs, the above equality will not hold true. thereefore y+z <> 0