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by GMATGuruNY » Sat Jun 01, 2013 3:12 pm
I believe that the problem 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 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 3:14 am, edited 1 time in total.
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by fcabanski » Sat Jun 01, 2013 11:26 pm
Here's an alternate answer. Mitch's is much nicer, but some people may not recall/realize they should use critical points.

Look at all the possibilities, find an answer, plug it in to check.

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

Use the two versions of the first abs value.

1. x+3-(4-x)=x-8 ---> x+3 -4 + x = x-8 ---> 2x-1=x-8--->x = -7 (Plug it into the original equation. It doesn't work. Discard it.)

2. 3-x -(4-x)=x-8 ---> 3-x-4+x=x-8 ---> -1 = x-8 ---> 7 = x (Doesn't work in the original equation. Discard it.)

Use the two versions of the second abs value.

3. x+3 - (x-4) = x-8 ---> x+3 -x + 4 = x-8 ---> 7 = x-8 ---> 15 = x (Works in the original equation. It's a solution.)

4. x+3 - (4-x) = x-8 ---> x+3 -4 +x = x-8 ---> 2x -1 = x-8 ---> x = -7 (Discard. Doesn't work)

Use the positive right side with each of the above four.

1. 2x-1=8-x ---> 3x = 9 ---> x=3 (Works in the original equation. Keep it.)

2. -1 = 8-x ---> 9 = x (Doesn't work in the original equation. Discard it.)


3. 7 = 8-x ---> 1 = x (Doesn't work. Discard.)

4. 2x -1 = 8-x ---> x = 3 (It works. Keep it.)

x=15 and x=3 are the solutions.
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by faraz_jeddah » Mon Jun 03, 2013 11:49 am
GMATGuruNY wrote:I believe that the problem 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 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.
Why do we flip the sign if the expression is negative? Is there a logic to it?
Also, which expression do we check ? The complete equation or just x+3 or 4-x or 8-x
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by faraz_jeddah » Mon Jun 03, 2013 11:52 am
Also Mitch you say

Since |x+3|<0 in this range, so we have to flip the signs in this expression.

How can an absolute value be negative?
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by fcabanski » Mon Jun 03, 2013 12:07 pm
Mitch, who provided the first solution, provided the compact, quick, best math solution. "Also, which expression do we check ? The complete equation or just x+3 or 4-x or 8-x " critical points tell you which equations to check.

I provided a slow, dirty, more difficult solution to show that even if you don't remember to set critical points, you can still solve absolute value problems.

"Why do we flip the sign if the expression is negative? Is there a logic to it?"

If there's a variable in an absolute value, the variable could be positive or negative.

|x| = x if x is already positive. For example, if x=3 then |x| = x because |3| = 3.

|x| = -x if x is negative. For example, if x=-3 then |x| = -x because |-3| = 3. The sign changed.

Because of the variable, we don't know the value of the statement inside the absolute value. If it's negative, then abs value flips the sign. If it's positive, abs value doesn't flip the sign.

For example, when x = 3 |x+2| = |5| = 5. The sign doesn't flip.

But if x = -3 |x+2| = |-3+2| = |-1| = 1. The sign flips.
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by fcabanski » Mon Jun 03, 2013 12:16 pm
"Also Mitch you say Since |x+3|<0 in this range, so we have to flip the signs in this expression. How can an absolute value be negative?"

The value is not negative, but it is opposite of the original value. We flip the sign because the value switched from negative to positive.

When x < -3, x+3 < 0 (it is negative). So |x+3| will be opposite of its original value.

For example: x=-10.

x+ 3 = -10+3 = -7

|-7| = 7. The sign flipped. 7 = -(-7)
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