If y [u]>[/u] 0, what is the value of x?

[justprashant]

Ben,

Can you explain why I cannot substitute. The above poster does so as well (Ron replied to that one). Confused here. Thanks.

Given:
y>=0

(1)
|x-3| >=y
|x-3| >= y >=0.

Therefore, |x-3|>=0.

i think i see what ben is getting at here. in a sense, you guys are both correct -- you're just talking past each other.

YOU are using the transitive property:
if a > b, and b > c, then a > c.
this is of course true.

BEN, however, is not talking so much about the inequalities themselves as he's talking about their solution sets.
for instance:
let's say that x > y and y > 10.
in this case,
* we do know that x > 10 (in other words, it's impossible that x > 10 is false); in that sense, you are correct.
* where you are going wrong here, though -- and what ben is trying to point out -- is that you are assuming that, given the above two inequalities, that x can automatically assume ANY value > 10. in other words, it seems that you are taking the two inequalities x > y and y > 10, together, as a guarantee that x can actually equal 10, or any greater number.
this isn't necessarily true, depending upon the value of y. for instance, if y is 15, then x is now restricted to numbers that are 15 and higher.

hope this makes more sense

Hi Ron,
It's still not clear. As per your sessions

| a | >= x should be read as

| a | >= x or | a | <= - x

Applying the same logic

1. |x-3| >=y and |x-3| <= -y
since y >= 0
|x-3| >= 0 or |x-3| <= -(0)
so x >=3 and x <=3

2. |x-3| <= - y
|x-3| <= -(0) or |x-3| >= 0
x <=3 or x >= 3

combining x = 3. where am I going wrong? Please advice

[RonPurewal]

2. |x-3| <= - y
|x-3| <= -(0) or |x-3| >= 0
x <=3 or x >= 3

these steps are invalid here, because (-y) is a negative number.
as a simpler analogy, consider the following inequalities:
|x - 3| < 4
|x - 3| < -4
the first of these can be solved with normal techniques (i.e., change it into -4 < x - 3 < 4).
the second can't -- it has no solutions at all, since an absolute value can't be negative.

if you still aren't seeing how this applies to the situation at hand, try numbers until you see what is happening.
i.e., go ahead and actually try to find specific values of "x" and "y" (with "y" positive, as required) that satisfy the equation |x - 3| < -y.
after you try enough values and observe what happens, (a) you'll realize that the equation can only work if x = 3 and y = 0, and (b) you'll see why.

[justprashant]

2. |x-3| <= - y
|x-3| <= -(0) or |x-3| >= 0
x <=3 or x >= 3

these steps are invalid here, because (-y) is a negative number.
as a simpler analogy, consider the following inequalities:
|x - 3| < 4
|x - 3| < -4
the first of these can be solved with normal techniques (i.e., change it into -4 < x - 3 < 4).
the second can't -- it has no solutions at all, since an absolute value can't be negative.

if you still aren't seeing how this applies to the situation at hand, try numbers until you see what is happening.
i.e., go ahead and actually try to find specific values of "x" and "y" (with "y" positive, as required) that satisfy the equation |x - 3| < -y.
after you try enough values and observe what happens, (a) you'll realize that the equation can only work if x = 3 and y = 0, and (b) you'll see why.

Thanks Ron. Still I' am unable to comprehend it.
|x - 3| <=-y
Before I look at the values of Y, I will write the equations

Step 1 :|x - 3| <= -y or |x - 3| >=y

Step 2: since absolute value cannot be negative,
|x - 3| = 0 Hence x =3

Case 2: |x - 3| >= 0
so x >= 3.

Please tell me where am I going wrong. Spent so much of time but still unclear.

[RonPurewal]

Before I look at the values of Y, I will write the equations

Step 1 :|x - 3| <= -y or |x - 3| >=y

this is a big mistake -- this kind of equation (in which an absolute value is less than a number) NEVER splits into "or".

if you are going to approach this algebraically, then all such equations become "and":
|blah| < A
becomes
blah < A and blah > -A, or
-A < blah < A.

so, here, that would be |x - 3| < -y and |x - 3| > y, or
y < |x - 3| < -y

since y is 0 or greater, the only possible way for this to work is y = 0. (if y is positive, then you get an impossible condition.) so, same conclusion.

Case 2: |x - 3| >= 0
so x >= 3.

i don't know where this is coming from.

[justprashant]

Before I look at the values of Y, I will write the equations

Step 1 :|x - 3| <= -y or |x - 3| >=y

this is a big mistake -- this kind of equation (in which an absolute value is less than a number) NEVER splits into "or".

if you are going to approach this algebraically, then all such equations become "and":
|blah| < A
becomes
blah < A and blah > -A, or
-A < blah < A.

so, here, that would be |x - 3| < -y and |x - 3| > y, or
y < |x - 3| < -y

since y is 0 or greater, the only possible way for this to work is y = 0. (if y is positive, then you get an impossible condition.) so, same conclusion.


Thanks ron. Now, I realised the mistake. I was using "OR" instead of AND.Can I say, when I have something like |x - 3| <= -y the equation transfers to AND instead of OR?

Case 2: |x - 3| >= 0
so x >= 3.

i don't know where this is coming from.

I used the above condition i.e |x - 3| >=y and used the value of Y as >=0

[tim]

at this point, your trail of work on this problem is becoming confusing. if you still have questions about this one, please summarize your work along with explanations of each step, and point out where you got stuck so we can see where exactly you need help on this one..

[davetzulin]

have a question on my approach, which is purely algebraic:

stmt 1

|x-3| >= y

case 1 (stuff in abs value is negative)
x-3 < 0 same as x < 3

-(x-3) >= y
-x + 3 >= y
but y >= 0, so by transitivity
-x + 3 >= 0
-x >= -3
x <= 3

** this is weird, how come i have x <=3 as a result when x-3 < 0 which means x < 3? I understand, yes, x = 3 is a possibility, but i'm not testing for those how did it appear?


case 2 (stuff in abs value is positive)
x-3 > 0 same as x > 3

x-3 >= y
but y >= 0, so again
x-3 >= 0
x >= 3

** again this is weird, yes i understand x = 3 is a possibility but I "limited" the case to x > 3


stmt 2

* I used the same approach as above and in case 1 and case 2 you end up with a contradictory result which means x-3 cannot be negative and x-3 cannot be positive, leaving the only possibility x-3 = 0, which works.

my question is really on stmt 1.

[RonPurewal]

dave, the problem can be seen in your labels here:

case 1 (stuff in abs value is negative)
...
case 2 (stuff in abs value is positive)

these labels aren't 100% accurate. specifically, your "case 1" covers all the possibilities in which the "stuff" is negative or zero, and your "case 2" covers all the possibilities in which the "stuff" is positive or zero. that explains why the equality is turning up in both cases: because it's true in both cases.

if you don't have a good grip on this, try some simpler equations: |x| = x
|x| = -x
the first of these is true when x > 0 (not just when x > 0), and the second is true when x < 0 (not just when x < 0). BOTH of them are true when x is zero.

[davetzulin]

dave, the problem can be seen in your labels here:

case 1 (stuff in abs value is negative)
...
case 2 (stuff in abs value is positive)

these labels aren't 100% accurate. specifically, your "case 1" covers all the possibilities in which the "stuff" is negative or zero, and your "case 2" covers all the possibilities in which the "stuff" is positive or zero. that explains why the equality is turning up in both cases: because it's true in both cases.

if you don't have a good grip on this, try some simpler equations: |x| = x
|x| = -x
the first of these is true when x > 0 (not just when x > 0), and the second is true when x < 0 (not just when x < 0). BOTH of them are true when x is zero.

wow thanks Ron. if you didn't list that rudimentary example it would have taken me a while to see that!

i think it is a mistake then to write |x| ,for example, into two cases (which i notice is a knee-jerk reaction in how most people do this algebraically)

x < 0
x > 0

it's actually a negative of the entire expression for any value of x. not the value of x being negative or positive.

[RonPurewal]

dave, the problem can be seen in your labels here:

case 1 (stuff in abs value is negative)
...
case 2 (stuff in abs value is positive)

these labels aren't 100% accurate. specifically, your "case 1" covers all the possibilities in which the "stuff" is negative or zero, and your "case 2" covers all the possibilities in which the "stuff" is positive or zero. that explains why the equality is turning up in both cases: because it's true in both cases.

if you don't have a good grip on this, try some simpler equations: |x| = x
|x| = -x
the first of these is true when x > 0 (not just when x > 0), and the second is true when x < 0 (not just when x < 0). BOTH of them are true when x is zero.

wow thanks Ron. if you didn't list that rudimentary example it would have taken me a while to see that!

i think it is a mistake then to write |x| ,for example, into two cases (which i notice is a knee-jerk reaction in how most people do this algebraically)

x < 0
x > 0

it's actually a negative of the entire expression for any value of x. not the value of x being negative or positive.

i don't know exactly what you mean in the last sentence.

in any case, you should learn to think of absolute value as a "washing machine" into which you can throw mathematical quantities. if they are dirty (negative) when you throw them in, they come out clean (positive). if they are already clean when you throw them in, they still come out clean.

[supratim7]

If y ≥ 0,
Rephrase: either y = 0 or y = "some positive"

what is the value of x?

1. |x - 3| ≥ y
Rephrase: Distance between x & 3 ≥ y

Combining (1) & y ≥ 0, Possible scenarios..
Distance between x & 3 = 0. So, x = 3
Distance between x & 3 > 0 (e.g. > 1). So, x = 5, x = 1.5, etc.
Distance between x & 3 = "some positive" (e.g. 1). So, x = 5, x = 1.5, etc.
Distance between x & 3 > "some positive" (e.g. > 1). So, x = 5, x = 1.5, etc.

INSUFFICIENT.

2. |x - 3| ≤ -y
Rephrase: Distance between x & 3 ≤ -y

Combining (2) & y ≥ 0, Possible scenarios..
Distance between x & 3 = -0 (i.e. = 0). So, x = 3
Distance between x & 3 < -0 (i.e. < 0). Impossible; distance always positive or 0.
Distance between x & 3 = -"some positive" (i.e. = "some negative"). Impossible; distance always positive or 0.
Distance between x & 3 < -"some positive" (i.e. < "some negative"). Impossible; distance always positive or 0.

SUFFICIENT: B

Surprised that Ron didn't refer his tried n tested way of rephrasing |x - y| = "distance between x & y". I gave it a try.. hope it's fine.

Many thanks | Supratim

[RonPurewal]

Yep, that works, too.

If you already have in mind the idea that "distance between 2 numbers" can't be negative, then it's unnecessary to break the second statement into all those cases"”since all the cases where -y represents a negative number are clearly impossible. But, if you don't have that insight right away, then that's the kind of breakdown you can (and probably should) do.

[supratim7]

If you already have in mind the idea that "distance between 2 numbers" can't be negative, then it's unnecessary to break the second statement into all those cases"”since all the cases where -y represents a negative number are clearly impossible.

Agree. I got that idea early on.. Didn't lay out all those cases where -y represents a negative number while solving. Just thought of laying out things clearly in a post..

Thank you for the reply :)

[jlucero]

Appreciate you keeping it clean for others. Not only does it help them out, but it hopefully helps clarify when certain cases do/don't work. Ultimately, the real test is about the most bang (correct answer choices) for your buck ( hard work).