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Complex Absolute Value Concept

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Complex Absolute Value Concept

by greenwich » Fri Jan 17, 2014 2:57 pm
|x+3| - |4-x| = |8+x|. How many solutions does the equation have?

I'm not quite sure if this type of multiple Absolute Values question will be asked in GMAT. Can someone walk me through the steps for this question?
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by Patrick_GMATFix » Fri Jan 17, 2014 10:56 pm
GMAT questions with 3 abs. value expressions are certainly rare.

This is an equation with no solution.

I would solve this by conceptualizing abs. value as a measure of distance. |a-b| is the distance from a to b on the number line. Just try it; |2-5| is the distance from 2 to 5. Absolute value of |2+5| = |2-(-5)| and is the distance from 2 to -5. |5|=|5-0| and is the distance from 5 to 0. Thinking of abs. value as a measure of distance can sometimes greatly simplify the problem.

Rewrite the equation as |x-(-3)| = |x-(-8)| + |x-4|. In English this equation says that "the distance between -3 and x is the sum of the distance between -8 and x and the distance between x and 4".

Is this possible?

Case 1: could x be between -8 and 4?
For all values of x between -8 and 4 (inclusive), the distance between -3 and x will vary from a minimum of 0 (if x=-3) to a maximum of 7 (if x=4). At the same time, the sum of the distance between -8 and x and the distance between x and 4 will always be 12 (if x is between -8 and 4, then this sum is equal to the distance from -8 to 4).
So Case 1 is not possible since the range 0 to 7 does not overlap with 12.


Case 2: could x be greater than 4?
For all values of x greater than 4, the distance from -3 to x is shorter than the distance from -8 to x, so the equation will not hold true (remember the equation tells us that distance from -3 to x equals the sum of distance from -8 to x and distance from x to 4).
So Case 2 is not possible.


Case 3: could x be less than -8?
For all values of x less than -8, the distance from -3 to x is shorter than the distance from 4 to x, so the equation will not hold true (for the equation to be true, distance from -3 to x must equal the sum of distance from 4 to x and distance from -8 to x).
So Case 3 is not possible.

Since x cannot be between -8 and 4 inclusive, cannot be greater than 4, and cannot be less then -8, there is no solution.

Hope that helped,
-Patrick
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by GMATGuruNY » Sat Jan 18, 2014 4:46 am
I believe that the question stem should read as follows:
|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 within the absolute values might be LESS than 0 -- unacceptable, since an absolute value must be nonnegative.
Implication:
If within a given range an expression will be less than 0, we must 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.
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by vinni.k » Wed Jan 29, 2014 9:42 am
Nice explanation. But Mitch how we can get to know that 4-x<0 and 8-x<0 for 3rd and 4th range. (confused on this part)

Regards
Vinni
GMATGuruNY wrote:
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 within the absolute values might be LESS than 0 -- unacceptable, since an absolute value must be nonnegative.
Implication:
If within a given range an expression will be less than 0, we must FLIP ITS SIGNS.

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.
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