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# If y>=0, What is the value of x?

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If y>=0, What is the value of x? [#permalink]

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21 Jan 2012, 08:57
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If y >= 0, What is the value of x?

(1) |x-3| >= y
(2) |x-3| <= -y

[Reveal] Spoiler:
I solved this way -

1) |x-3| >= y

x -3 >=y
x >= y + 3

or

3 - x >=y
x <= 3 - y

so y + 3 <= x <= 3 - y

if y is positive then the above condition is not possible as sum of two positives cant be less than difference of two positives. the other possibility is that y is zero and x = 3. hence sufficient?

2) |x-3| <= -y

x -3 <= -y
x <=3-y

or

3-x <=-y
x>=y+3

so 3-y >= x >= y+3, again since y >=0 , only possibility is that y = 0. hence x = 3. sufficient.

But the OA is different.
[Reveal] Spoiler: OA

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21 Jan 2012, 09:23
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Apex231 wrote:
if y >= 0, What is the value of x?
1) |x-3| >= y
2) |x-3| <= -y

I solved this way -

1) |x-3| >= y

x -3 >=y
x >= y + 3

or

3 - x >=y
x <= 3 - y

so y + 3 <= x <= 3 - y

if y is positive then the above condition is not possible as sum of two positives cant be less than difference of two positives. the other possibility is that y is zero and x = 3. hence sufficient?

2) |x-3| <= -y

x -3 <= -y
x <=3-y

or

3-x <=-y
x>=y+3

so 3-y >= x >= y+3, again since y >=0 , only possibility is that y = 0. hence x = 3. sufficient.

But the OA is different.

If $$y\geq{0}$$, what is the value of [b]x?[/b]

(1) $$|x - 3|\geq{y}$$. As given that $$y$$ is non negative value then $$|x - 3|$$ is more than (or equal to) some non negative value, (we could say the same ourselves as absolute value in our case ($$|x - 3|$$) is never negative). So we can not determine single numerical value of $$x$$. Not sufficient.

Or another way: to check $$|x - 3|\geq{y}\geq{0}$$ is sufficient or not just plug numbers:
A. $$x=5$$, $$y=1>0$$, and B. $$x=8$$, $$y=2>0$$: you'll see that both fits in $$|x - 3|>=y$$, $$y\geq{0}$$.

Or another way:
$$|x - 3|\geq{y}$$ means that:

$$x - 3\geq{y}\geq{0}$$ when $$x-3>0$$ --> $$x>3$$

OR (not and)
$$-x+3\geq{y}\geq{0}$$ when $$x-3<0$$ --> $$x<3$$

Generally speaking $$|x - 3|\geq{y}\geq{0}$$ means that $$|x - 3|$$, an absolute value, is not negative. So, there's no way you'll get a unique value for $$x$$. INSUFFICIENT.

(2) $$|x-3|\leq{-y}$$. Now, as $$|x-3|$$ is never negative ($$0\leq{|x-3|}$$) then $$0\leq{-y}$$ --> $$y\leq{0}$$ BUT stem says that $$y\geq{0}$$ thus $$y=0$$. $$|x-3|\leq{0}$$ --> $$|x-3|=0=y$$ (as absolute value, in our case |x-3|, can not be less than zero) --> $$x-3=0$$ --> $$x=3$$. SUFFICIENT

In other words:
$$-y$$ is zero or less, and the absolute value ($$|x-3|$$) must be at zero or below this value. But absolute value (in this case $$|x-3|$$) can not be less than zero, so it must be $$0$$.

There is a following problem with your solution:
If $$x<3$$ --> $$-(x-3)\geq{y}$$ --> $$3-y\geq{x}$$;
OR:
If $$x\geq{3}$$ --> $$(x-3)\geq{y}$$ --> $$x\geq{3+y}$$;

But you can not combine these inequalities and write: $$3+y\leq{x}\leq{3-y}$$ as they are OR scenarios not AND scenarios (meaning that depending on the value of x we'll have either the first one or the second one).

Also discussed here: if-y-geq-0-what-is-the-value-of-x-1-x-3-geq-y-91640.html

Hope it helps..
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21 Jan 2012, 09:50
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Bunuel wrote:
But you can not combine these inequalities and write: $$3+y\leq{x}\leq{3-y}$$ as they are OR scenarios not AND scenarios (meaning that depending on the value of x we'll have either the first one or the second one).

Thanks Bunuel..realize my mistake now...great explanation as always...

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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

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02 Jul 2013, 00:54
What difficulty level would this be closer to 650 or above that? BTW Great Explanation as always!
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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

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02 Jul 2013, 00:59
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Expert's post
fozzzy wrote:
What difficulty level would this be closer to 650 or above that? BTW Great Explanation as always!

The question involves 2 variables, inequalities and modulus, plus it's quite tricky, so I'd say it's above 650, more like 700.
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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

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08 Jul 2013, 17:14
If y >= 0, What is the value of x?

(1) |x-3| >= y
if y>=0 then |x-3| must be greater than y.
|x-3| >= y
x>=3:
|x-3| >= y
x-3 >=y
x>=y+3
There is no need to test for a negative case here as we are told that |x-3| is greater than or equal to a positive number. Still, we don't know anything about x except that it must be greater than or equal to 3.
INSUFFICIENT

(2) |x-3| <= -y
This means that x-3 is negative, so:
-(x-3)<=-y
3-x<=-y
-(3-x)<=y

Bunuel, I get your explanation, but I have to be honest, I don't think I would have any idea on the test to employ the explanation you got as opposed to the one I started with. Can you shed some light on to how you utilized the particular solution you used (i.e. what signs should I look for to know to solve this problem this way?)

Thanks!

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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

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08 Jul 2013, 18:38
WholeLottaLove wrote:
If y >= 0, What is the value of x?

(1) |x-3| >= y
if y>=0 then |x-3| must be greater than y.
|x-3| >= y
x>=3:
|x-3| >= y
x-3 >=y
x>=y+3
There is no need to test for a negative case here as we are told that |x-3| is greater than or equal to a positive number. Still, we don't know anything about x except that it must be greater than or equal to 3.
INSUFFICIENT

We're not told |x-3| is greater than or equal to a positive number. We're told that y is either 0 or a positive number and that |x-3| >= y. So |x-3| can equal 0. But since |anything|>=0 this doesn't tell us anything more than we already know. In other words, the absolute value of any number is not negative and 1 simply states that |x-3| is not negative.

Quote:
(2) |x-3| <= -y
This means that x-3 is negative, so:
-(x-3)<=-y
3-x<=-y
-(3-x)<=y

|x-3|<=-y does not mean that x-3 is negative. The key to the problem is to remember that the absolute value of anything is not negative. The difference in 2 is that, since we know y>=0, we know that -y <=0. But since 2 states that |x-3|<=-y and it is a fact that the absolute value of anything is not negative, -y=0 and so y=0. Then it's clear that |x-3|=0 and so x=3. Anytime absolute value shows up the trick may likely be remembering the basic fact that absolute value is not negative but that not negative is not the same as positive. There is a possibility of 0.

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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

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08 Jul 2013, 20:31
Hmmmm...I am still quite lost on your explanation for 2)

Observer wrote:
WholeLottaLove wrote:
If y >= 0, What is the value of x?

(1) |x-3| >= y
if y>=0 then |x-3| must be greater than y.
|x-3| >= y
x>=3:
|x-3| >= y
x-3 >=y
x>=y+3
There is no need to test for a negative case here as we are told that |x-3| is greater than or equal to a positive number. Still, we don't know anything about x except that it must be greater than or equal to 3.
INSUFFICIENT

We're not told |x-3| is greater than or equal to a positive number. We're told that y is either 0 or a positive number and that |x-3| >= y. So |x-3| can equal 0. But since |anything|>=0 this doesn't tell us anything more than we already know. In other words, the absolute value of any number is not negative and 1 simply states that |x-3| is not negative.

Quote:
(2) |x-3| <= -y
This means that x-3 is negative, so:
-(x-3)<=-y
3-x<=-y
-(3-x)<=y

|x-3|<=-y does not mean that x-3 is negative. The key to the problem is to remember that the absolute value of anything is not negative. The difference in 2 is that, since we know y>=0, we know that -y <=0. But since 2 states that |x-3|<=-y and it is a fact that the absolute value of anything is not negative, -y=0 and so y=0. Then it's clear that |x-3|=0 and so x=3. Anytime absolute value shows up the trick may likely be remembering the basic fact that absolute value is not negative but that not negative is not the same as positive. There is a possibility of 0.

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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

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09 Jul 2013, 02:03
WholeLottaLove wrote:
(2) |x-3| <= -y
This means that x-3 is negative, so:
-(x-3)<=-y
3-x<=-y
-(3-x)<=y

Dear WholeLottaLove

Your point "This means that x-3 is negative" is incorrect.

General theory is: $$|X| <= a$$, ==> $$-a <= X <= a$$
So, X does not have to be negative. X is between -a and a.
For example: $$|x -3| <= 9$$ ==> $$-9 <= (x -3) <=9$$

The KEY point is absolute value cannot be negative, thus the smallest value of $$|x - 3|$$ is 0.
Because $$-y <= 0$$ ==> $$|x-3|$$ must be 0 ==> x = 3.

Hope it helps.
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Re: If y>=0, What is the value of x? (1) |x-3|>=y (2) |x-3|<=-y [#permalink]

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09 Jul 2013, 14:05

First number line: y>=0
Second: From properties of absolute value, |x-3|>=0 is true no matter what x is.
Third: (2) tells us that |x-3|<=-y. We know nothing about y other than y>=0. Therefore |x-3|<=-y only tells us that |x-3|<=0.
Fourth: Since |x-3|<=0 AND |x-3|>=0, we now know the exact value of |x-3|. In the diagram it is the point on the number line that is both <=0 and >=0. The only value for which that is true is 0. So |x-3|=0. From there you know that x-3=0 and so x=3.

WholeLottaLove wrote:
Hmmmm...I am still quite lost on your explanation for 2)

Observer wrote:
WholeLottaLove wrote:
If y >= 0, What is the value of x?

(1) |x-3| >= y
if y>=0 then |x-3| must be greater than y.
|x-3| >= y
x>=3:
|x-3| >= y
x-3 >=y
x>=y+3
There is no need to test for a negative case here as we are told that |x-3| is greater than or equal to a positive number. Still, we don't know anything about x except that it must be greater than or equal to 3.
INSUFFICIENT

We're not told |x-3| is greater than or equal to a positive number. We're told that y is either 0 or a positive number and that |x-3| >= y. So |x-3| can equal 0. But since |anything|>=0 this doesn't tell us anything more than we already know. In other words, the absolute value of any number is not negative and 1 simply states that |x-3| is not negative.

Quote:
(2) |x-3| <= -y
This means that x-3 is negative, so:
-(x-3)<=-y
3-x<=-y
-(3-x)<=y

|x-3|<=-y does not mean that x-3 is negative. The key to the problem is to remember that the absolute value of anything is not negative. The difference in 2 is that, since we know y>=0, we know that -y <=0. But since 2 states that |x-3|<=-y and it is a fact that the absolute value of anything is not negative, -y=0 and so y=0. Then it's clear that |x-3|=0 and so x=3. Anytime absolute value shows up the trick may likely be remembering the basic fact that absolute value is not negative but that not negative is not the same as positive. There is a possibility of 0.

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Re: If y>=0, What is the value of x? [#permalink]

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10 Jul 2013, 13:43
If y >= 0, What is the value of x?

(1) |x-3| >= y
We are told that |x-3| ≥ y but all we know is that y is a positive # greater than or equal to zero. Therefore, all we know is that |x-3| is greater than or equal to zero and x could be an infinite number of possibilities.

(2) |x-3| <= -y

(I am having real difficulty understanding the rationale for (2) could someone explain it to me? (preferably like they would to a 5th grader )

Thanks!

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Re: If y>=0, What is the value of x? [#permalink]

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10 Jul 2013, 14:58
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WholeLottaLove wrote:
If y >= 0, What is the value of x?

(1) |x-3| >= y
We are told that |x-3| ≥ y but all we know is that y is a positive # greater than or equal to zero. Therefore, all we know is that |x-3| is greater than or equal to zero and x could be an infinite number of possibilities.

(2) |x-3| <= -y

(I am having real difficulty understanding the rationale for (2) could someone explain it to me? (preferably like they would to a 5th grader )

Thanks!

Hi WholeLottaLove

The important property concerning absolute value inequalities is:
|a| <= b <--> -b <= a <=b
[a is in the middle of -b and b, inclusive]

Apply to statement (2)
|x - 3| <= -y
<--> -(-y) <= (x-3) <= -y,
<--> y <= (x-3) <= -y

We know y is 0 or positive, -y is 0 or negative.
Because there is not any number that is both negative and positive.Thus, there is ONLY one number that is both >= y AND <= -y. That is zero.
Therefore, (x-3) must be zero. ==> x = 3

Hope it's clear.
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Re: If y>=0, What is the value of x? [#permalink]

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11 Jul 2013, 11:18
pqhai wrote:
WholeLottaLove wrote:
If y >= 0, What is the value of x?

(1) |x-3| >= y
We are told that |x-3| ≥ y but all we know is that y is a positive # greater than or equal to zero. Therefore, all we know is that |x-3| is greater than or equal to zero and x could be an infinite number of possibilities.

(2) |x-3| <= -y

(I am having real difficulty understanding the rationale for (2) could someone explain it to me? (preferably like they would to a 5th grader )

Thanks!

Hi WholeLottaLove

The important property concerning absolute value inequalities is:
|a| <= b <--> -b <= a <=b
[a is in the middle of -b and b, inclusive]

Apply to statement (2)
|x - 3| <= -y
<--> -(-y) <= (x-3) <= -y,
<--> y <= (x-3) <= -y

We know y is 0 or positive, -y is 0 or negative.
Because there is not any number that is both negative and positive.Thus, there is ONLY one number that is both >= y AND <= -y. That is zero.
Therefore, (x-3) must be zero. ==> x = 3

Hope it's clear.

Just to be clear (because sometimes I can be a bit dense!) we know that y is zero or negative because it is less than or equal to (x-3) which is less than or equal to -y, correct? In other words, y has to be less than or equal to zero because it is less than or equal to a negative number. The intersection of -y and y>= 0 is zero, right?

Also, do you know of any similar problems that I could practice on?

Thanks!

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Re: If y>=0, What is the value of x? [#permalink]

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11 Jul 2013, 13:23
1
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WholeLottaLove wrote:
pqhai wrote:
WholeLottaLove wrote:
If y >= 0, What is the value of x?

(1) |x-3| >= y
We are told that |x-3| ≥ y but all we know is that y is a positive # greater than or equal to zero. Therefore, all we know is that |x-3| is greater than or equal to zero and x could be an infinite number of possibilities.

(2) |x-3| <= -y

(I am having real difficulty understanding the rationale for (2) could someone explain it to me? (preferably like they would to a 5th grader )

Thanks!

Hi WholeLottaLove

The important property concerning absolute value inequalities is:
|a| <= b <--> -b <= a <=b
[a is in the middle of -b and b, inclusive]

Apply to statement (2)
|x - 3| <= -y
<--> -(-y) <= (x-3) <= -y,
<--> y <= (x-3) <= -y

We know y is 0 or positive, -y is 0 or negative.
Because there is not any number that is both negative and positive.Thus, there is ONLY one number that is both >= y AND <= -y. That is zero.
Therefore, (x-3) must be zero. ==> x = 3

Hope it's clear.

Just to be clear (because sometimes I can be a bit dense!) we know that y is zero or negative because it is less than or equal to (x-3) which is less than or equal to -y, correct? In other words, y has to be less than or equal to zero because it is less than or equal to a negative number. The intersection of -y and y>= 0 is zero, right?

Also, do you know of any similar problems that I could practice on?

Thanks!

Hi WholeLottaLove

Yes, you're correct. The key point is -y <= 0, and according to the inequality: y <= (x-3) <= -y
So y <= -y. But y >= 0. Thus, y must be zero so -y = y = 0.

I would like to clarify a bit. My post above is to explain why the intersection must be zero in the case |A| <= zero or negative number.
The KEY of this question is -y <= 0 (a zero or negative number), and the smallest value |x-3| (absolute value) is zero, so (x-3) must be zero.
If you see a "regular" question like: |x-3| <= 9 (a positive number). The range of (x-3) is from -9 to 9.

Hope it's more clear.

PS: If I know more similar questions, I will pm you.
_________________

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Re: If y>=0, What is the value of x? [#permalink]

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11 Jul 2013, 14:00
Thank you for the explanation. I think I understand it better!

<---------------------------------------------------------y<=0
-10 0 10
y>=0--------------------------------------------------------------->

Along those lines?

Hi WholeLottaLove

Yes, you're correct. The key point is -y <= 0, and according to the inequality: y <= (x-3) <= -y
So y <= -y. But y >= 0. Thus, y must be zero so -y = y = 0.

I would like to clarify a bit. My post above is to explain why the intersection must be zero in the case |A| <= zero or negative number.
The KEY of this question is -y <= 0 (a zero or negative number), and the smallest value |x-3| (absolute value) is zero, so (x-3) must be zero.
If you see a "regular" question like: |x-3| <= 9 (a positive number). The range of (x-3) is from -9 to 9.

Hope it's more clear.

PS: If I know more similar questions, I will pm you.[/quote]

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Re: If y>=0, What is the value of x? [#permalink]

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11 Jul 2013, 14:05
So, (as you said) if y were 9 then -y = -9

|x-3| can be at the very least, 0 (as it is an absolute value) so if we are told that |x-3| is less than negative y, then y can be at the very most zero (because obviously -y cannot be a positive number) so if y is a negative number and it is greater than something that can be at least zero, the only option is for y to be zero (|0| <= -0)

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Re: If y>=0, What is the value of x? [#permalink]

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11 Jul 2013, 14:14
If y >= 0, What is the value of x?

(1) |x-3| >= y

If y>=0 then |x-3| >= to any positive number. X cannot be determined.
INSUFFICIENT

(2) |x-3| <= -y

So, |x-3| can never be negative which means that it cannot be less than a negative number. For |x-3| <= -y to be true, y must be negative:
|x-3| <= -y
|x-3| <= -(-y)
|x-3| <= (y)

It is possible for |x-3| to be less than or equal to y.

So, y is less than or equal to zero but we are also told that y is greater than or equal to zero, therefore the intersection is at 0.

I think I understand!!

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Re: If y>=0, What is the value of x? [#permalink]

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11 Jul 2013, 15:02
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WholeLottaLove wrote:
If y >= 0, What is the value of x?

(1) |x-3| >= y

If y>=0 then |x-3| >= to any positive number. X cannot be determined.
INSUFFICIENT

(2) |x-3| <= -y

So, |x-3| can never be negative which means that it cannot be less than a negative number. For |x-3| <= -y to be true, y must be negative:
YOU"RE CORRECT. THAT'S THE KEY.

|x-3| <= -y
|x-3| <= -(-y)
|x-3| <= (y)

It is possible for |x-3| to be less than or equal to y.

So, y is less than or equal to zero but we are also told that y is greater than or equal to zero, therefore the intersection is at 0.

I think I understand!!

Cheer!
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