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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

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

Re: Abs DS from OG [#permalink]
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 [ 7.25 KiB | Viewed 4940 times ]

Last edited by Andr359 on 15 Jan 2007, 08:48, edited 1 time in total.

Re: Abs DS from OG [#permalink]
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?

Re: Abs DS from OG [#permalink]
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.

Re: Abs DS from OG [#permalink]
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. _________________

In a Normal Distribution, only the Average'Stand Out'

Re: If x and y are integers and y=|x+3|+|4-x| does y=7? 1)x<4 [#permalink]
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. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

My last interview took place at the Johnson School of Management at Cornell University. Since it was my final interview, I had my answers to the general interview questions...