Problem Solving

solve for possible value of x ; /x + 1/ + /x - 3/ = 6.

only 3 possible scenarios:
1. when x < -1
(x + 1) is negative in this region and (x - 3) is also negative, so the equation becomes: -(x + 1) - (x - 3) = 6, which simplifies to x = -2
With complex absolute value questions, the solution (x = -2) must be checked against the range for which the equation holds true (x < -1). In this case there is no conflict (since -2 is less than -1) so this is an actual solution.

2. when -1 < x < 3 (x + 1) is positive in this region and (x - 3) is negative, so the equation becomes:(x + 1) - (x - 3) = 6, which simplifies to 4 = 6, i.e. mathematical jibberish!
This means that there is no solution for the equation in this range.

3. when x > 3 (x + 1) is positive in this region and (x - 3) is also positive, so the equation becomes: (x + 1) + (x - 3) = 6, which simplifies to x = 4
This solution (x = 4) check out since it is in the range for which the equation hold true (x > 3).

Therefore, there are two potential solutions to this absolute value equation, x = -2 and 4.

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/x - 4 / > 8

Scenario 1: when x - 4 > 0 (i.e. when x > 4), x - 4 > 8, which simplifies to x > 12.
Scenario 2: when x - 4 < 0 (i.e. when x < 4), -(x - 4) > 8 which simplifies to x < -4
The solution is x > 12 OR x < -4.
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Is x > 0 ?
(1) /x - 3/ < 5
(2) /x + 2/ > 5

The first statement can be solved using the method described above.
Scenario 1: When x > 3, x - 3 < 5, which can be simplified to x < 8
Scenario 2: When x < 3, -(x - 3) < 5 or x > -2
Statement (1) can be simplified as -2 < x < 8, which is NOT sufficient to answer the question "is x > 0?"

The second statement can be solved in a similar manner.
Scenario 1: When x > -2, x + 2 > 5 or x > 3.
Scenario 2: When x < -2, -(x + 2) > 5 or x < -7.
Statement (2) can be simplified as x > 3 or x < -7, which is NOT sufficient to answer the question "is x > 0?"

Since both statements are true (remember this is always the case in Data Sufficiency), x must be greater than 3 and less than 8. This is the only overlapping region between the ranges from statements (1) and (2). All of the values between 3 and 8 are positive so TOGETHER the statements are SUFFICIENT and the answer is C.

Concerning the first example, when there's one variable only, i thought you needed to consider only two real cases : one in which neither expression changes sign, and another in which one expression changes sign. So I got x=4 and x= -1/2 as a result. Please explain this issue a little further.

Concerning the first example, when there's one variable only, i thought you needed to consider only two real cases : one in which neither expression changes sign, and another in which one expression changes sign. So I got x=4 and x= -1/2 as a result. Please explain this issue a little further.

vikram4689 wrote:solve for possible value of x ; /x + 1/ + /x - 3/ = 6.

Here's an alternate approach that -- with practice -- will be very fast.

|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 distance 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.

Thanks! That's a great tip.
However, isn't this specific to two sets of absolute values.

What if |x-4| = 8 was the case. Here, the distance will be -x--------4--- So to make the distance = 8, x must increase by 4 units on the negative axis and 8 on the positive axis.

so x can be -4 or x can be 12.

Also, what if it was:
|x-a| + |x-b| + |x-c|... n = d.. Is there a more generic approach to this?

Thanks a lot!

GMATGuruNY wrote:

vikram4689 wrote:solve for possible value of x ; /x + 1/ + /x - 3/ = 6.

Here's an alternate approach that -- with practice -- will be very fast.

}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 distances will increase by TWO UNITS.
Thus:
To increase the sum of the 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 distance 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.

prepp wrote:Thanks! That's a great tip.
However, isn't this specific to two sets of absolute values.

What if |x-4| = 8 was the case.

|x-4| = 8 means the distance between x and 4 is 8 units.
In other words, x is either 8 units to the left of 4 or 8 units to the right of 4:
-4<-------4-------->12
Thus, x=-4 or x=12.

Also, what if it was:
|x-a| + |x-b| + |x-c|... n = d.. Is there a more generic approach to this?

This sort of complexity is beyond the scope of the GMAT.

Awesome! Thanks for that.

GMATGuruNY wrote:

prepp wrote:Thanks! That's a great tip.
However, isn't this specific to two sets of absolute values.

What if |x-4| = 8 was the case.

|x-4| = 8 means the distance between x and 4 is 8 units.
In other words, x is either 8 units to the left of 4 or 8 units to the right of 4:
-4<-------4-------->12
Thus, x=-4 or x=12.

Also, what if it was:
|x-a| + |x-b| + |x-c|... n = d.. Is there a more generic approach to this?

This sort of complexity is beyond the scope of the GMAT.