absolute value problem using open modulus

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topspin360: Master | Next Rank: 500 Posts; Posts: 103; Joined: Sat Jun 02, 2012 9:46 pm; Thanked: 1 times

absolute value problem using open modulus

by topspin360 » Fri Aug 10, 2012 7:08 pm

can anyone show how to do this problem using open modulus?

|X+3|-|4-x|=|8+x|

How many solutions does the equations have?

so 3 roots could be -3, 4, -8?

Can I just pick numbers in between these roots and see if the equation holds up? That's how I would tackle this problem if it was inequality.. does anything change now that it's an equation?

Also, in the following setup, why do we set two terms to negative and leave one as is:
if x<-8, -(x+3)-(4-x) = -(8+x)...

Thanks

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Source: Beat The GMAT — Problem Solving | Fri Aug 10, 2012 7:08 pm

Pharo: Senior | Next Rank: 100 Posts; Posts: 61; Joined: Mon Jan 18, 2010 11:09 pm; Thanked: 8 times

by Pharo » Sat Aug 11, 2012 5:14 am

What do you mean by open modulus?

The way I would solve this problem is by evaluating how I would calculate the absolute values. I would note down the points at which the "opening of the absolute functions" changes.

|x+3| : changes when x < -3;
|4-x| : changes when x > 4;
|8+x| : changes when x < -8

So you have to evaluate the equality in the cases below:

Case1: if x < -8;
|x+3| would "open" as -(x+3); say x =-9, x+3 = -9+3 = -6 Since outcome is negative, it opens with negative multiplier
|4-x| would "open" as (4-x); say x = -9, 4-x = 4 - (-9) = 13 Since outcome is positive, it opens with positive multiplier
|8+x| would "open" as -(8+x); you can use the above logic for this as well.
Putting all together:
-(x+3) - (4-x) = -(8+x)
-x-3-4+x=-8-x
x=-1; this is not a valid answer since we started with the assumption that x<-8.

Case2: if -8<=x<-3
|x+3| would "open" as -(x+3)
|4-x| would "open" as (4-x)
|8+x| would "open" as (8+x)
Putting all together:
-(x+3) - (4-x) = (8+x)
-x-3-4+x=8+x
-7=8+x
x=-15; this is not a valid answer since we started with the assumption that -8<=x<-3

Case3: if -3<=x<4;
|x+3| would "open" as (x+3)
|4-x| would "open" as (4-x)
|8+x| would "open" as (8+x)
Putting all together:
(x+3)-(4-x) = (8+x)
x+3-4+x=8+x
2x-1=8+x
x=9; this is not a valid answer since we started with the assumption that -3<=x<=4

Case4: if 4<=x
|x+3| would "open" as (x+3)
|4-x| would "open" as -(4-x)
|8+x| would "open" as (8+x)
Putting it all together:
(x+3)- (-(4-x))=(8+x)
x+3+4-x=8+x
x=-1; this is not a valid answer since we started with the assumption that 4<=x
Result:
The answer is 0 since we could not find any solutions.

By the way, in answer to your first question about picking roots and plugging them in: No, you cannot. This is not an inequality problem. You have to solve the x.

The answer of your second question is in the Case1.

Hope this helps

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

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by GMATGuruNY » Sat Aug 11, 2012 6:00 am

Did you mean the following?

|x+3|-|4-x|=|8-x|.
How many solutions does the equation have?

The critical points are -3, 4 and 8.
These are the values where the expressions within the absolute values are equal to 0.
To each side of these critical points, some of the expressions in the absolute values might be negative while others are positive.
If an expression will be negative within a given range, we need to flip its signs.

x<-3:
Since x+3<0 in this range, so we have to flip the signs in this expression.
-x-3 - (4-x) = 8-x
-7 = 8-x
x=15.
Since only values such that x<-3 are valid in this range, x=15 is not a valid solution here.

-3<x<4:
Since none of the expressions is less than 0 in this range, no signs need to be flipped.
x+3 - (4-x) = 8-x
2x-1 = 8-x
x=3.

4<x<8:
Since 4-x<0 in this range, we have to flip the signs in this expression.
x+3 - (-4+x) = 8-x
7 = 8-x
x=1.
Since only values such that 4<x<8 are valid in this range, x=1 is not a valid solution here.

x>8:
Since 4-x<0 and 8-x<0 in this range, we have to flip the signs in these expressions.
x+3 - (-4+x) = -8+x
7 = -8+x
x=15.
(We can save time by recognizing that flipping the signs of one or more expressions is the equivalent of flipping the signs of the OTHER expressions.
Thus, flipping the signs of 4-x and 8-x will yield the same result as did flipping the signs of x+3.
Thus, the solution in both cases will be the same: x=15, which is valid for x>8.)

The solutions of the equation are x=3 and x=15.

Last edited by GMATGuruNY on Wed Oct 30, 2013 3:20 am, edited 3 times in total.

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topspin360: Master | Next Rank: 500 Posts; Posts: 103; Joined: Sat Jun 02, 2012 9:46 pm; Thanked: 1 times

by topspin360 » Sat Aug 11, 2012 8:29 am

Thank you Pharo and Mitch, great explanations. I'm confusing absolute values with inequalities b/c I'm trying to use number line approach to solve both type of problems.. this explains how absolute values should be tackled.

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Pharo: Senior | Next Rank: 100 Posts; Posts: 61; Joined: Mon Jan 18, 2010 11:09 pm; Thanked: 8 times

by Pharo » Sun Aug 12, 2012 3:03 am

No worries

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optimist: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Wed Jun 06, 2012 9:53 am; Thanked: 1 times

by optimist » Wed Aug 15, 2012 2:48 am

GMATGuruNY wrote:Did you mean the following?

|x+3|-|4-x|=|8-x|.
How many solutions does the equation have?
The critical points are -3, 4 and 8.
These are the values where the expressions within the absolute values are equal to 0.
To each side of these critical points, some of the expressions in the absolute values might be negative while others are positive.
If an expression will be negative with the given range, we need to flip its signs.

x<-3:
Since |x+3|<0 in this range, so we have to flip the signs in this expression.
-x-3 - (4-x) = 8-x
-7 = 8-x
x=15.
Since only values such that x<-3 are valid in this range, x=15 is not a valid solution here.

-3<x<4:
Since none of the expressions is less than 0 in this range, no signs need to be flipped.
x+3 - (4-x) = 8-x
2x-1 = 8-x
x=3.

4<x<8:
Since |4-x|<0 in this range, we have to flip the signs in this expression.
x+3 - (-4+x) = 8-x
7 = 8-x
x=1.
Since only values such that 4<x<8 are valid in this range, x=1 is not a valid solution here.

x>8:
Since |4-x|<0 and |8-x|<0 in this range, we have to flip the signs in these expressions.
x+3 - (-4+x) = -8+x
7 = -8+x
x=15.
(We can save time by recognizing that flipping the signs of one or more expressions is the equivalent of flipping the signs of the OTHER expressions.
Thus, flipping the signs of |4-x| and |8-x| will yield the same result as did flipping the signs of |x+3|.
Thus, the solution in both cases will be the same: x=15, which is valid for x>8.)

The solutions of the equation are x=3 and x=15.

Hi GMATGuru

The question originally posted has |8+x|, but you have posted the solution with |8-x|. With the former, the question has no solution but one can find a solution to the latter. Is the question posted by you, the correct question?