Manhattan method produces a wrong ANSWER-Please clear

[dddanny2006]

Is x ≥ 0?

1) x^2 = 9x

2) |x| = -x

Statement 1 is absolutely sufficient.So the answer is A or D.When it comes to statment 2,this is my way of solving it and it contradicts with the real answer.

|x|=-x => x=+(-x)=-x,this is true when x>0, and x=-(-x),this is true when x<0
Therefore statement 2 gives us 2 conditions x=x and x=-x.Now since x=-x when x>0 ,x can never be equal to -x when x>0.So I dont know what I can call this condition.Now,we also have x=x when x<0,so -5=-5,-6=-6,these dont satisfy the condition in the question stem and hence No.ThereforeStatement 2 gives us 2 conditions of which one condition gives us a _________(I dont know what to call it because it cant be tested due to x>0) and the other condition gives us a No(-1,-2... are not greater than or equal to 0).When we have a statement that has 2 sub statements,and one of those sub statements(x=-x and x=x subjected to x>0 for the former and x<0 for the latter) cant be tested,but the other one gives us a No.What bearing does this have on that statement?Is it a Yes or a No for Statement 2?Please help me solve this on the basis of the Manhattan article here.

In which way is my understanding of the question wrong?

I have followed this technique from here--
http://www.manhattangmat.com/strategy-s ... -value.cfm

Absolute value expressions start to become difficult when variable expressions are placed inside the bars. For example, /x/. Upon a cursory examination, the expression /x/ seems like it should be equal to x. Since there is no sign in front of the x, the absolute value bars should be able to be removed without jeopardizing the "guarantee of positive." What this line of reasoning fails to account for, however, is that x itself could be negative! When dealing with absolute value expressions that contain variables, two scenarios must be considered: (1) the scenario whereby the expression inside the bars is positive and (2) the scenario whereby the expression inside the bars is negative.

In this example, for scenario (1) if x > 0, the expression /x/ can simply be represented as x; for scenario (2) if x < 0, the expression /x/ must be represented as (-x). Notice that in the negative scenario, we don't simply remove the absolute value bars. We remove the absolute value bars and negate the entire expression within.

Let's look at a more complicated example: the expression /x - 3/. As always, we must consider both the positive and negative scenarios. When is the expression inside the absolute value bars positive? Not simply when x > 0, but when x - 3 > 0 or when x > 3. Likewise the expression will be negative when x < 3.

To recap, the two scenarios are:
(1) /x - 3/ can be rewritten as x - 3 when x > 3
(2) /x - 3/ can be rewritten as -(x - 3) or 3 - x when x < 3

One more for the road: /3x + y/.
(1) /3x + y/ can be rewritten as 3x + y when 3x + y > 0
(2) /3x + y/ can be rewritten as -(3x + y) when 3x + y < 0

[RonPurewal]

Please don't post the same question more than once. If the same question is posted more than once by the same user, we'll wait to answer it until after we've responded to all other posts.
Thanks.

Discussion here:
post92961.html#p92961

[dddanny2006]

Ok,sorry about that.Point taken

Is x ≥ 0?

1) x^2 = 9x

2) |x| = -x

Statement 1 is absolutely sufficient.So the answer is A or D.When it comes to statment 2,this is my way of solving it and it contradicts with the real answer.

|x|=-x => x=+(-x)=-x,this is true when x>0, and x=-(-x),this is true when x<0
Therefore statement 2 gives us 2 conditions x=x and x=-x.Now since x=-x when x>0 ,x can never be equal to -x when x>0.So I dont know what I can call this condition.Now,we also have x=x when x<0,so -5=-5,-6=-6,these dont satisfy the condition in the question stem and hence No.ThereforeStatement 2 gives us 2 conditions of which one condition gives us a _________(I dont know what to call it because it cant be tested due to x>0) and the other condition gives us a No(-1,-2... are not greater than or equal to 0).When we have a statement that has 2 sub statements,and one of those sub statements(x=-x and x=x subjected to x>0 for the former and x<0 for the latter) cant be tested,but the other one gives us a No.What bearing does this have on that statement?Is it a Yes or a No for Statement 2?Please help me solve this on the basis of the Manhattan article here.

In which way is my understanding of the question wrong?

I have followed this technique from here--
http://www.manhattangmat.com/strategy-s ... -value.cfm

Absolute value expressions start to become difficult when variable expressions are placed inside the bars. For example, /x/. Upon a cursory examination, the expression /x/ seems like it should be equal to x. Since there is no sign in front of the x, the absolute value bars should be able to be removed without jeopardizing the "guarantee of positive." What this line of reasoning fails to account for, however, is that x itself could be negative! When dealing with absolute value expressions that contain variables, two scenarios must be considered: (1) the scenario whereby the expression inside the bars is positive and (2) the scenario whereby the expression inside the bars is negative.

In this example, for scenario (1) if x > 0, the expression /x/ can simply be represented as x; for scenario (2) if x < 0, the expression /x/ must be represented as (-x). Notice that in the negative scenario, we don't simply remove the absolute value bars. We remove the absolute value bars and negate the entire expression within.

Let's look at a more complicated example: the expression /x - 3/. As always, we must consider both the positive and negative scenarios. When is the expression inside the absolute value bars positive? Not simply when x > 0, but when x - 3 > 0 or when x > 3. Likewise the expression will be negative when x < 3.

To recap, the two scenarios are:
(1) /x - 3/ can be rewritten as x - 3 when x > 3
(2) /x - 3/ can be rewritten as -(x - 3) or 3 - x when x < 3

One more for the road: /3x + y/.
(1) /3x + y/ can be rewritten as 3x + y when 3x + y > 0
(2) /3x + y/ can be rewritten as -(3x + y) when 3x + y < 0

[RonPurewal]

No worries