Absolute value - testing numbers or algebra

Many absolute value problems can be solved as DISTANCE problems.

|x-y|= the distance between x and y.
|x+y| = |x-(-y)| = the distance between x and -y.

Thus, |x + 1| + |x - 3| = 6 implies the following:
(the distance between x and -1) + (the distance between x and 3) = 6.
In other words, the SUM OF THE TWO DISTANCES = 6.

-1--------x----------3
The distance between -1 and 3 is 4.
If x is BETWEEN -1 and 3, the sum of the two distances will be 4.
For the sum of the two distances to be 6, x must be to the left of the lower endpoint (-1) or to the right of the upper endpoint (3).

For every unit x moves beyond either endpoint, the SUM of the two distances will increase by TWO UNITS.
The reason is that EACH DISTANCE is affected by the movement of x.
Since there are two distances, the effect is TWICE AS GREAT.
Thus:
To increase the sum of the two distances by k units, x must be (1/2)k units to the left of the lower endpoint or (1/2)k units to the right of the upper endpoint.

Since the sum here must increase by 2 units, x must be 1 unit to the left of -1 or 1 unit to the right of 3:
x=-2<----(-1)---------------(3)----->x=4

Thus, there are two valid solutions for x:
x=-2 and x=4.

Here's another example:

|x-2| + |x+3| = 13.

-3---------------2
The distance between -3 and 2 is 5.
To increase the sum of the two red distances by 8 to 13, x must be 4 units to the left of -3 or 4 units to the right of 2:
x=-7<----(-3)---------------(2)----->x=6

Thus, there are two valid solutions for x:
x=-7 and x=6.

infiniti007 wrote:What is the value of integer x?

1.) |1-x| - |x+1| = 0
2.) |7-x| + |3-x| = 10

Statement 1: |1-x| - |x+1| = 0
|1-x| = |x+1|
The distance between 1 and x is equal to the distance between x and -1.
In other words, x is HALFWAY between 1 and -1.
Thus, x=0.
SUFFICIENT.

Statement 2: |7-x| + |3-x| = 10
Rephrased:
|x-7| + |x-3| = 10.

3---------------7
The distance between 3 and 7 is 4.
To increase the sum of the two red distances by 6 to 10, x must be 3 units to the left of 3 or 3 units to the right of 7:
x=0<----(3)---------------(2)----->x=10

Thus, there are two valid solutions for x:
x=0 and x=10.
INSUFFICIENT.

The correct answer is A.

Thanks Mitch - really like the approach. Do you know if this approach can be used in general for most of these problem types?

As one other example, could it be applied when the data sufficiency statements provide inequalities?

Example:

If y = |x + 7| + |2 - x|, is y = 9?

1.) x < 2
2.) x > -7

infiniti007 wrote: If y = |x + 7| + |2 - x|, is y = 9?

1.) x < 2
2.) x > -7

|a-b| = the DISTANCE between a and b.
|a+b| = |a-(-b)| = the DISTANCE between a and -b.

Thus:
|x+7| = the distance between x and -7.
|2-x| = the distance between 2 and x.
y = the SUM of these two distances.

Question stem rephrased:
Is the sum of the two distances equal to 9?

The distance between -7 and 2 is 9.

Thus, if x is BETWEEN these two endpoints, then the sum of the two distances will be EQUAL TO 9:
-7 <--- |x+7| ---> x <---|2-x|---> 2.
Here, |x+7| + |2-x| = the distance between -7 and 2 = 9.

By extension, if x is BEYOND either endpoint -- if x is to the left of -7 or to the right of 2 -- then the sum of the two distances will be GREATER THAN 9.

Statement 1: x < 2
If x=1, then x is between -7 and 2.
If x=-10, then x is to the left of -7.
INSUFFICIENT.

Statement 2: x > -7
If x=1, then x is between -7 and 2.
If x=10, then x is to the right of 2.
INSUFFICIENT.

Statements combined:
-7 < x < 2.
SUFFICIENT.

The correct answer is C.

More practice:
https://www.beatthegmat.com/algebra-ineq ... 31389.html

GMATGuruNY wrote:Many absolute value problems can be solved as DISTANCE problems.

|x-y|= the distance between x and y.
|x+y| = |x-(-y)| = the distance between x and -y.

Thus, |x + 1| + |x - 3| = 6 implies the following:
(the distance between x and -1) + (the distance between x and 3) = 6.
In other words, the SUM OF THE TWO DISTANCES = 6.

-1--------x----------3
The distance between -1 and 3 is 4.
If x is BETWEEN -1 and 3, the sum of the two distances will be 4.
For the sum of the two distances to be 6, x must be to the left of the lower endpoint (-1) or to the right of the upper endpoint (3).

For every unit x moves beyond either endpoint, the SUM of the two distances will increase by TWO UNITS.
The reason is that EACH DISTANCE is affected by the movement of x.
Since there are two distances, the effect is TWICE AS GREAT.
Thus:
To increase the sum of the two distances by k units, x must be (1/2)k units to the left of the lower endpoint or (1/2)k units to the right of the upper endpoint.

Since the sum here must increase by 2 units, x must be 1 unit to the left of -1 or 1 unit to the right of 3:
x=-2<----(-1)---------------(3)----->x=4

Thus, there are two valid solutions for x:
x=-2 and x=4.

Here's another example:

|x-2| + |x+3| = 13.

-3---------------2
The distance between -3 and 2 is 5.
To increase the sum of the two red distances by 8 to 13, x must be 4 units to the left of -3 or 4 units to the right of 2:
x=-7<----(-3)---------------(2)----->x=6

Thus, there are two valid solutions for x:
x=-7 and x=6.

infiniti007 wrote:What is the value of integer x?

1.) |1-x| - |x+1| = 0
2.) |7-x| + |3-x| = 10

Statement 1: |1-x| - |x+1| = 0
|1-x| = |x+1|
The distance between 1 and x is equal to the distance between x and -1.
In other words, x is HALFWAY between 1 and -1.
Thus, x=0.
SUFFICIENT.

Statement 2: |7-x| + |3-x| = 10
Rephrased:
|x-7| + |x-3| = 10.

3---------------7
The distance between 3 and 7 is 4.
To increase the sum of the two red distances by 6 to 10, x must be 3 units to the left of 3 or 3 units to the right of 7:
x=0<----(3)---------------(2)----->x=10

Thus, there are two valid solutions for x:
x=0 and x=10.
INSUFFICIENT.

The correct answer is A.

HI Mitch.

WOW & Great Explanation

I want to know in your example what if I change the equation to

|x + 1| - |x - 3| = 6

Does it make any difference

Thanks

Mo2men wrote:

GMATGuruNY wrote:Many absolute value problems can be solved as DISTANCE problems.

|x-y|= the distance between x and y.
|x+y| = |x-(-y)| = the distance between x and -y.

Thus, |x + 1| + |x - 3| = 6 implies the following:
(the distance between x and -1) + (the distance between x and 3) = 6.
In other words, the SUM OF THE TWO DISTANCES = 6.

-1--------x----------3
The distance between -1 and 3 is 4.
If x is BETWEEN -1 and 3, the sum of the two distances will be 4.
For the sum of the two distances to be 6, x must be to the left of the lower endpoint (-1) or to the right of the upper endpoint (3).

For every unit x moves beyond either endpoint, the SUM of the two distances will increase by TWO UNITS.
The reason is that EACH DISTANCE is affected by the movement of x.
Since there are two distances, the effect is TWICE AS GREAT.
Thus:
To increase the sum of the two distances by k units, x must be (1/2)k units to the left of the lower endpoint or (1/2)k units to the right of the upper endpoint.

Since the sum here must increase by 2 units, x must be 1 unit to the left of -1 or 1 unit to the right of 3:
x=-2<----(-1)---------------(3)----->x=4

Thus, there are two valid solutions for x:
x=-2 and x=4.

Here's another example:

|x-2| + |x+3| = 13.

-3---------------2
The distance between -3 and 2 is 5.
To increase the sum of the two red distances by 8 to 13, x must be 4 units to the left of -3 or 4 units to the right of 2:
x=-7<----(-3)---------------(2)----->x=6

Thus, there are two valid solutions for x:
x=-7 and x=6.

infiniti007 wrote:What is the value of integer x?

1.) |1-x| - |x+1| = 0
2.) |7-x| + |3-x| = 10

Statement 1: |1-x| - |x+1| = 0
|1-x| = |x+1|
The distance between 1 and x is equal to the distance between x and -1.
In other words, x is HALFWAY between 1 and -1.
Thus, x=0.
SUFFICIENT.

Statement 2: |7-x| + |3-x| = 10
Rephrased:
|x-7| + |x-3| = 10.

3---------------7
The distance between 3 and 7 is 4.
To increase the sum of the two red distances by 6 to 10, x must be 3 units to the left of 3 or 3 units to the right of 7:
x=0<----(3)---------------(2)----->x=10

Thus, there are two valid solutions for x:
x=0 and x=10.
INSUFFICIENT.

The correct answer is A.

HI Mitch.

WOW & Great Explanation

I want to know in your example what if I change the equation to

|x + 1| - |x - 3| = 6

Does it make any difference

Thanks

no. it doesn't. it just means you are subtracting the distance between x and 3 from x and -1, instead of adding.

in the above case, there is no solution, because the distance between 3 (from |x-3|) and -1 (from |x+1|) is 4. the moment x is placed beyond either of these end points, the "difference" between the distances remains 4 all the time. Hence, there is no solution of x that could give a difference of 6.

Mo2men wrote:HI Mitch.

WOW & Great Explanation

I want to know in your example what if I change the equation to

|x + 1| - |x - 3| = 6

Does it make any difference

Thanks

As nrmlvrm has noted, the equation above has no solution.
For a discussion of this sort of problem, check my third post here:
https://www.beatthegmat.com/absolute-val ... tml#762724