What is the value of integer x?

What is the value of integer x?


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


Can anyone help in how do we open the modules sign for 1 and 2?

this is how i solved it but its time consuming
1. |x-1| - |x+1| = 0
-(x-1) - (x+1) = 0
therefore x= 0
(x-1) + (x+1) = 0
therefore x = 0

any alternate technique to solve this problem

Make sure to format your question in such a way that puts your analysis under "spoiler".

As for your question,

Both statements can be solved in a similar way,

Statement 1, |1-x|-|x+1|=0 , now as |a-x|=|x-a| , you get |x-1|-|x+1|=0 ---> |x-1|=|x+1| ----> \(\pm (x-1) = \pm (x+1)\) , you only get 2 possible cases giving you x=0 for both cases. Hence this statement is sufficient.

Statement 2, |7 - x| + |3 + x| = 10, using the same principle as shown in statement 1, you get the equation as \(\pm (x-7) \pm (x+3) = 10\), giving you 2 possible cases again but giving you x=-3 and x=7. Thus this statement is not sufficient.

Hope this helps.

I am no longer active on this forum.

I didn't square, since I didn't know how to tackle this question rather than try the possible options, and although I selected A as the answer I later check and found something interesting:
both positive
both negative
1st positive second negative
1st negative second positive:

1. |1 - x| - |x + 1| = 0
++
1-x-x+1=0=> 0=2x=>x=0
--
-1+x+x+1=>2x=0 => x=0
-+
-1+x-x+1 => 0=0 - doesn't help us much, since there is no way to solve it. or x can take any value.
+-
1-x+x+1 => 2=0 - this is not a solution.

x
2. |7 - x| + |3 + x| = 10

only 2 options tested, which yielded 2 different values for x, thus I dismissed B and D.

For such questions the most straightforward method to solve is to realize that |x-a| with a = constant is the distance of x from 'a' and |a-x| = |x-a|

Thus |1-x| = |x-1| --> distance of x from 1,

Similarly for |x+1|, |x-7| and |x+3|.

Do remember that x = integer (given information).

Per statement 1, |1 - x| - |x + 1| = 0 ---> |x-1| - |x + 1| = 0 ---> |x-1| = |x + 1| ---> distance of integer x is same from -1 and 1 ---> x can only take 0 as the possible value. Unique value of x ---> sufficient statement.

Per statement 2, |7 - x| + |3 + x| = 10 --> |x-7| + |3 + x| = 10 ---> distance of x from 7 + distance of x from -3 = 10 units. This is satisfied by all integer values of x between -3 and 7. Consider x = 4 , distance from 7 = 3 , distance of x from -3 = 7 , total = 10.

But if x = 0, distance from 7 = 7 and distance of x from -3 = 3 , total = 10 units.

Thus this statement provides multiple values of x ---> not sufficient.

A is thus the correct answer.

Hope this helps.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

What is the value of integer x?

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

When it comes to absolute value, it indicates a distance between two dots. That is, it is a distance between |x-y|=x and y in this question.
In the original condition, there is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
For 1), from |1-x|=|x-(-1)|, since a distance from x to 1 is as same as a distance from x to -1, x=0 is derived, which is unique and sufficient.
For 2), from |7-x|+|x-(-3)|=10, the sum of a distance between x and 7, and a distance between x and -3 is 10. That is, all values are possible in -3<=x<=7, which is not unique and not sufficient. Therefore, the answer is A.


 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

Bunuel:
How do I deal with second case -
|7-x| + |3+x| = 10
Now four possibilities can exist
Case 1: 7-x>0 and 3+x>0
-3 < x < 7
On solving
10=10

Case 2: 7-x<0 and 3+x<0
4=10
Dump this!

Case 3: 7-x>0 and 3+x<0
x<7 and x<-3
or x<-3
On solving we get ; x=-3

Case 4: 7-x<0 and 3+x>0
x > -3 and x >7
or x>7
On solving
x = 7

Bunuel: Can you please suggest how do I conclude this.

\(|3 + x| + |7 - x| = 10\)

2 key points at -3 and 7.

When \(x < -3\), then \(3 + x\) is negative and \(7 - x\) is positive, thus \(|3 + x| + |7 - x| = 10\) becomes \(-(3 + x) + (7 - x) = 10\) --> \(x=-3\). Discard since -3 is not in the range;

When \(-3 \leq x \leq 7\), then \(3 + x\) is positive and \(7 - x\) is positive, thus \(|3 + x| + |7 - x| = 10\) becomes \((3 + x) + (7 - x) = 10\) --> \(10=10\). Since this is true, then this means that ANY x from the given range satisfies the equation;

When \(x > 7\), then \(3 + x\) is positive and \(7 - x\) is negative, thus \(|3 + x| + |7 - x| = 10\) becomes \((3 + x) - (7 - x) = 10\) --> \(x=7\). Discard since 7 is not in the range.

Therefore, \(|3 + x| + |7 - x| = 10\) is true for any x where \(-3 \leq x \leq 7\).

Hope it's clear.

Statement 1 can be phrased as: |1−x|=|x+1|. Trial and error shows that the two will only be equal if x=0. Accordingly, Statement 1 is sufficient.

Statement 2, on the other hand, works for many integers. Simply trying and finding two integers that work is enough to know that multiple answers are possible: If x=3, then the statement holds: |7−3|=4; |3+3|=6; and 4+6=10. And if x=0, then the statement still holds: 7+3=10. Therefore, statement 2 is not sufficient, and again the correct answer is A.

Hi Bunuel and chetan2u

i have a conceptual query on such questions..

whenever we have the question of the form in stmt 2; we usually open the mod by using a combination of ++, +-, -+ right?

Hi Bunuel, I had a question for statement (2).

For values of x < -3, I get:

(7 - x) - (3 + x) = 10

Solving this, I get x = -3

Now I am confused. We started with assuming that x < -3, but got the solution that x = -3.

What should we deduce out of this?

I explained this question in detail here (including your doubt): https://gmatclub.com/forum/what-is-the- ... l#p1741469

Thanks Bunuel. So basically, we started with assuming that x < -3, but got the solution that x = -3

Since x = -3 is not in the range of x < -3, we basically discard x = -3 as a solution.

Bunuel regarding the first step , in which you square both values, do the elements within an absolute phrase have to fulfill some requirements in order for us to be able to square both parts and act as the | | were just a parenthesis, or do we square it because the value of an absolute phrase is always positive so we can square the absolute value whenever we want to.

(1) \(|1 - x| - |x + 1| = 0\)

\(|1 - x| = |x + 1|\)

\(1 - x = x + 1 or 1 -x = -(x + 1)\)

\(x = 0\)

SUFFICIENT.

(2) \(|7 - x| + |3 + x| = 10\)

Taking a few seconds to analyze this statement, we see that multiple values work for this equation. For example, if x = 0 or if x = 7. INSUFFICIENT.

Answer is A.

Bunuel i think there is typo in your post

_____________________
Edited. Thank you!