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If x and y are integers and y=|x+3|+|4-x| does y=7?

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If x and y are integers and y=|x+3|+|4-x| does y=7? [#permalink] New post 13 Jan 2007, 11:41
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If x and y are integers and y=|x+3|+|4-x| does y=7?

(1) x<4
(2) x>-3

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-integers-and-y-x-3-4-x-does-y-equals-101766.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Oct 2012, 07:14, edited 2 times in total.
Renamed the topic and edited the question.
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 [#permalink] New post 13 Jan 2007, 11:54
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(C) for me :)

y=|x+3|+|4-x|

The special points are : x = -3 and x = 4

o If x < -3, y is simplified to:
y = -(x+3) + 4-x
<=> y = 1 - 2*x

o If -3 <= x <= 4, y is simplified to:
y = (x+3) + 4-x
<=> y = 7

o If x > 4, y is simplified to:
y = (x+3) + -(4-x)
<=> y = 2*x - 1

The question is thus asking where x is.

Stat 1
x<4 : we have seen that if x < -3, y = 1 - 2*x > 7. Also, if -3 <= x < 4, then y =7.

INSUFF.

Stat 2
x > -3 : It's same problem.

o If -3 <= x < 4, then y =7
o If x > 4, then y = 2*x - 1

INSUFF.

Both (1) & (2)
We are in only one domain, -3 < x < 4, and we have so y = 7.

SUFF.
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 [#permalink] New post 13 Jan 2007, 13:10
Fig

WOW!! I read your explanation for about 15 minutes to understand it... Nothing against the way you have written... But my basics are weak and reading your explanation and understanding it took some time!!

So the key in putting the values for x is find a break point i.e. At what point y will not be equal to 7. In this case x < -3 and x> 4 y is not equal to 7.


The 2nd thing (to understand and freshen up on my basics) is that the moment value of x < -3 and x > 4, you removed the equation in between the mod signs (given below) and changed the sign and solved the equation.

o If x < -3, y is simplified to:
y = -(x+3) + 4-x
<=> y = 1 - 2*x

o If x > 4, y is simplified to:
y = (x+3) + -(4-x)
<=> y = 2*x - 1
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 [#permalink] New post 13 Jan 2007, 14:05
Quote:
The special points are : x = -3 and x = 4


How to figure those spl. points?
Should I just have in mind that the numbers that add up to zero are the spl. points?

y=|x+3|+|4-x|

Here, |-3+3|+|4-(-3)|=|0|+|7|=7
and |4+3|+|4-4|=|7|+|0|=7

Thanks Fig[/quote]
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 [#permalink] New post 13 Jan 2007, 14:11
Sumithra wrote:
Quote:
The special points are : x = -3 and x = 4


How to figure those spl. points?
Should I just have in mind that the numbers that add up to zero are the spl. points?

y=|x+3|+|4-x|

Here, |-3+3|+|4-(-3)|=|0|+|7|=7
and |4+3|+|4-4|=|7|+|0|=7

Thanks Fig


Yes... :) We need to look for each value of x that makes flip the sign in each absolute value. :)

For instances
o in |x - a| + |x - b| + |x - c| + |x - d| >>> the special points are a, b, c, and d.
o in |x - a|*|x - b| or |(x - a)(x - b)| >>> the special points are a and b
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 [#permalink] New post 13 Jan 2007, 14:34
Thank you. Good to know. Better than trying few values.
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 [#permalink] New post 13 Jan 2007, 15:25
Thanks Fig!! Appreciate it! Making more sense now!
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 [#permalink] New post 13 Jan 2007, 15:34
U are welcome guys :)
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Re: Abs DS from OG [#permalink] New post 14 Jan 2007, 19:25
Sumithra wrote:
If x and y are integers and y=|x+3|+|4-x| does y=7?

(1)x<4
(2)x>-3

Any basic idea to be kept in mind instead of trying few values?


By plotting the equation on the xy plane one can also arrive at the answer. For the range [-3 , 4]: y = x+3 + 4-x = 7. An essential thing to remember is that, if 2 lines with slopes m and -m (same absolute value, opposite signs) are added, the result is a horizontal line; in this case, the line y = 7. Hope the diagram is illustrative.
Attachments

abs val.jpg
abs val.jpg [ 7.25 KiB | Viewed 4107 times ]


Last edited by Andr359 on 15 Jan 2007, 08:48, edited 1 time in total.
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Re: Abs DS from OG [#permalink] New post 15 Jan 2007, 17:27
I think the simplest approach would be that since the question has ranges, it needs upper and lower boundary. and 1 & 2 together have upper and lower boundary in this case.

if 2) was x <-3, then answer for this would be E.


Sumithra wrote:
If x and y are integers and y=|x+3|+|4-x| does y=7?

1)x<4
2)x>-3

Any basic idea to be kept in mind instead of trying few values?
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 [#permalink] New post 29 Aug 2007, 05:09
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One other way to solve it is as follows.

For y=7
y=|x+3|+|4-x| = x+3 + 4-x = 7
i.e. |x+3| = x+3 and |4-x| = 4-x. This is only possible when both mod values are +ve.

i.e. x+3 >0 implies x>-3
and 4-x > 0 implies -x > -4 implies x < 4

So y=7 for all values of x where -3 < x < 4. Choice C gives this value.
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Re: Abs DS from OG [#permalink] New post 15 Jun 2009, 11:32
Since we are talking about absolute values with a + sign between them its pretty obvious that 1) and 2) can never be sufficient by themselves.

What is left is to test if -3<x<4. I did this by plugging in three numbers (just to be sure) in that range. It took me under 60 seconds to solve this one using that method.
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Re: Abs DS from OG [#permalink] New post 14 May 2010, 19:06
@ Fig
I've browsed through most of the problems on Absolute Value, and IMHO, you are the most facile & astute solver of questions on Modulus there is.
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Re: If x and y are integers and y=|x+3|+|4-x| does y=7? 1)x<4 [#permalink] New post 21 Oct 2012, 04:36
Could any one explain me how negative sign is used across :) ..BTW i am novice with absolute.

If x < -3, y is simplified to:
y = - :roll: (x+3) + 4-x
<=> y = 1 - 2*x

o If -3 <= x <= 4, y is simplified to:
y = (x+3) + 4-x
<=> y = 7

o If x > 4, y is simplified to:
y = (x+3) + - :roll: (4-x)
<=> y = 2*x - 1

If -3 <= x <= 4, y is simplified to:
y = (x+3) + 4-x
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Re: If x and y are integers and y=|x+3|+|4-x| does y=7? 1)x<4 [#permalink] New post 21 Oct 2012, 05:40
Sumithra wrote:
If x and y are integers and y=|x+3|+|4-x| does y=7?

1)x<4
2)x>-3

Any basic idea to be kept in mind instead of trying few values?


Use the meaning of absolute value: |x - a| is the distance between x and a on the number line.

|4 - x| = |x - 4| is the distance between x and 4.
|x + 3| = |x - (-3)| is the distance between x and -3.

Since the distance between -3 and 4 is 7 (=4 - (-3)), y = 7 for every x between -3 and 4.
Draw the number line, it is easy to understand. Also, we can immediately deduce that neither (1) nor (2) alone is sufficient.

Therefore, answer C.
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Re: If x and y are integers and y=|x+3|+|4-x| does y=7? [#permalink] New post 23 Oct 2012, 07:14
Expert's post
Sumithra wrote:
If x and y are integers and y=|x+3|+|4-x| does y=7?

(1) x<4
(2) x>-3

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-integers-and-y-x-3-4-x-does-y-equals-101766.html


OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-integers-and-y-x-3-4-x-does-y-equals-101766.html
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Re: If x and y are integers and y=|x+3|+|4-x| does y=7?   [#permalink] 23 Oct 2012, 07:14
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