Difference between A and B is taken to be A - B in GMAT (At some other places, we take difference to mean Greater - Smaller)
Distance from a point is always positive.

For any point to the left of -5, say x = -6,
Distance from -5 = 1
Distance from 5 = 11

Difference between "distance from -5" and "distance from 5" = 1 - 11 = -10

Yes, if x = -4.5, then y = -9. We need integer values of y which we get. This is to show that y will take values between -10 and 10. That, x could be a decimal (which is acceptable) but we might still get an integer value for y.
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For this question, it's useful to know that |a - b| represents the DISTANCE from point a to point b on the number line.
For example, |3 - 10| = the DISTANCE from 3 to 10 on the number line.
Since |3 - 10| = |-7| = 7, we know that 7 is the distance from 3 to 10 on the number line

So, in this case, |x - 5| = the distance from x to 5
Likewise, since x + 5 = x - (-5), we know that |x + 5| = |x - (-5)|
So, |x + 5| = the distance from x to -5

The equation y = |x + 5| - |x - 5| has two critical points. These are x-values that MINIMIZE the value of |x + 5| and MINIMIZE the value of |x - 5|
First, |x + 5| is minimized when x = -5. That is, when x = -5, |x + 5| = |(-5) + 5| = |0| = 0
Next, |x - 5| is minimized when x = 5. That is, when x = 5, |x - 5| = |5 - 5| = |0| = 0

Let's add these critical points ( x = 5 and x = -5) to the number line.


Notice that these critical points divide the number line into 3 regions.
Let's see what happens to the value of y when x lies in each region.

Let's start with region 1

When x lies in region 1 (e.g., x = -7), notice that the blue bar represents the value of |x - 5|
When x lies in region 1 (e.g., x = -7), notice that the red bar represents the value of |x + 5| (aka |x - (-5)|
Since the distance between -5 and 5 is 10, we can see that the length of the blue bar must be 10 units LONGER than the red bar
So, it must be true that |x + 5| - |x - 5| = -10

Now let's generalize.
For ANY value of x in region 1, the length of the blue bar will be 10 units LONGER than the red bar.
So, for ANY value of x in region 1, |x + 5| - |x - 5| = -10

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Now let's see what's going on in region 3

When x lies in region 3 (e.g., x = 8), notice that the blue bar represents the value of |x - 5|
When x lies in region 3 (e.g., x = 8), notice that the red bar represents the value of |x + 5| (aka |x - (-5)|
Since the distance between -5 and 5 is 10, we can see that the length of the red bar must be 10 units LONGER than the blue bar
So, it must be true that |x + 5| - |x - 5| = 10

To generalize, we can say that, for ANY value of x in region 3, |x + 5| - |x - 5| = 10

IMPORTANT ASIDE: At this point, we've shown that |x + 5| - |x - 5| can equal 10 and -10
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Now onto region 2
Let's see what happens when x = -4

If x = -4, then |x + 5| (the red bar) = 1 and |x - 5| = 9
So, |x + 5| - |x - 5| = 1 - 9 = -8

Now let's see what happens when x = 3.5
[img]https://i.imgur.com/1SyK8Ii.png[/img
If x = 3.5, then |x + 5| = 8.5 and |x - 5| = 1.5
So, |x + 5| - |x - 5| = 8.5 - 1.5 = 7
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CONCLUSION: At this point, we've shown that |x + 5| - |x - 5| can equal 10, -10, and all values BETWEEN 10 and -10

There are 21 INTEGERS, from -10 to 10 inclusive.
So, y can have 21 different integer values

Answer: E

Cheers,
Brent
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\(?\,\,\,:\,\,\,\# \,\,\,{\rm{integer}}\,\,{\rm{values}}\,\,{\rm{for}}\,\,\,y = \left| {x + 5} \right| - \left| {x - 5} \right|\)


\(\left( {\rm{i}} \right)\,\,\,x \le - 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = - \left( {x + 5} \right) - \left( {5 - x} \right) = - 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,1\,\,{\rm{integer}}\)


\(\left( {{\rm{ii}}} \right)\,\,\, - 5 < x \le 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \left( {x + 5} \right) - \left( {5 - x} \right) = 2x\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{other}}\,\,10 \cdot 2\,\,{\rm{integers}}\,\,\,\left( * \right)\,\,\,\)

\(\left( * \right)\,\,\,x\,\,{\mathop{\rm int}} \,\,\,\left( { - 4 \le x \le 5\,\,\, \Rightarrow \,\,\,10\,\,{\rm{options}}} \right)\,\,\,\,{\rm{or}}\,\,\,\,\,\left\{ \matrix{
\,x \ne {\mathop{\rm int}} \hfill \cr
\,x = {{{\mathop{\rm int}} } \over 2} \hfill \cr} \right.\,\,\,\,\,\,\,\,\left( { - 9 \le \mathop{\rm int}\,{\rm{odd}}\,\, \le 9\,\,\,\,\, \Rightarrow \,\,\,10\,\,{\rm{options}}} \right)\)


\(\left( {{\rm{iii}}} \right)\,\,\,x > 5\,\,\,\,\, \Rightarrow \,\,\,\,\,y = \left( {x + 5} \right) - \left( {x - 5} \right) = 10\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{already}}\,\,{\rm{obtained}}\,\,{\rm{in}}\,\,\left( {{\rm{ii}}} \right)\)



\(? = 1 + 20 = 21\)



This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Exactly, Bunuel!

This gives us (now explicitly) an immediate alternate solution:





There are 21 integers between -10 and 10, both of them included (see blue interval).

Regards,
Fabio.
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If y=|x+5|−|x−5| then y can take how many integer values?
We need to find integer values of y. x can be an integer or non-integer.


if you enter in x as -5,-4.5,-4,-3.5,-3,-2.5,-2,-1.5,-1,-0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5 in y=|x+5|−|x−5| then you will get 21 different values of y. Beyond -5 and 5 you will get integer values same as the ones you get by entering the above 21 numbers.

Hence 21.

For Y = | X+5 | - | X - 5 |, we can identify critical points as -5 and 5.
So X must be in following ranges. X < -5 or -5 <= X < 5 or X > 5

1. Considering X < -5, we can modify modulus as
Y = -(X + 5) + (X - 5)
= -X -5 + X -5
Y = - 10...….(1 integer solution of Y for X < -5)

2. Considering X >= 5, we can modify modulus as
Y = (X +5 ) - (X - 5)
= X + 5 - X + 5
Y = 10...……(1 integer solution of Y for X >= 5)

3. Considering 3rd and last possible range of X. -5 <= X < 5,
Y = (X + 5) + (X -5)
Y = 2X...….Thus X can take any value between -5 till 4.9. But need integer values of Y.
Y= 2X can give integer values of Y only when X is an integer or X is a fraction with .5 at the end. So the value X can take are ( -5, -4.5, -4, -3.5, -3, -2.5, -2, -1.5, -1, -0.5, 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5) so total 20 values. But X = -5 will give Y to be -10. We have already considered Y = -10 in 1st range. hence in this range we get 19 possible integer values of Y.

Thus total possible integer values of Y are 1 + 1 + 19 = 21

hence (E)

Given: \(y=|x+5|-|x-5|\),

Asked: \(y\) can take how many integer values?

When x = 5; y = 10
when x=-5; y = -10
When x <-5; y= -10
When x>5; y =10
-5<=x<=5; -10<=y<=10; 21 integer values

IMO E
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Hello Bunuel,
Can you please explain how did you obtain the range.
I am getting for Mod (X+5), for positive solution, (X+5)>=0, this implies X>=-5 and for negative solution X+5<0, this gives me range X<-5. And for Mod(X-5) , for positive solution , (X-5)>=0, this gives me X>=5 and (X-5)<0 gives me X<5.

Dear Bunuel,

I get a bit confused when it comes to selecting the ranges for x

For example, when you considered the first range
x<= -5
|x+5| would be -x - 5 if x<-5 but if x =-5 then |x+5| = x + 5 (should not change signs since it is equal to 0) and in that case it would be,
x+5 + x - 5 = 2x

Could you tell me where i am making a mistake? and help me in how to select the values for ranges?

I am also confused by the range selection in Bunuel's solution. If we look at Gmat math club book , if I interpret the solutions correctly, it is recommended to start with x<(lowest range number) check for range (lowest range number)<=x<(highest range number), and finally x>=(highest range number). It is not quite clear how Bunuel decided 2nd and 3rd ranges either.

JeffTargetTestPrep , ScottTargetTestPrep Could you please help. I learnt from TTP how to solve for y. But I'm not sure how to use x>5... etc.
I got 3 values, y=10, y=2x and y=-10 and i figured it has to be E but i'm not sure if this is the right way to do it? Could you please clarify? Thank you very much.

Upon looking at the equation that's given to us, we notice it contains two absolute value expressions: |x + 5| and |x - 5|. Recall that an absolute value expression is evaluated with respect to the sign of the expression, so if z < 0, then |z| = -z and if z > 0, then |z| = z. That's why we are interested in the signs of the expressions x + 5 and x - 5.
Let's begin with x + 5. It is not hard to see that x + 5 is positive if x > -5 and negative if x < -5. Similarly, x - 5 is positive if x > 5 and negative if x < 5. This gives us three possibilities: 1) x < -5 (so that both x + 5 and x - 5 are negative), 2) -5 < x < 5 (so that x + 5 is positive, but x - 5 is negative), 3) x > 5 (so that both x + 5 and x - 5 are positive).

In the first case, since both expressions are negative, |x + 5| = -(x + 5) and |x - 5| = -(x - 5). Substituting for the equation y = |x + 5| - |x - 5|, we get:

y = -(x + 5) - [-(x - 5)] = -x - 5 + x - 5 = -10.

So, one possible value for y is -10.

In the second case, we have |x + 5| = x + 5 and |x - 5| = -(x - 5). Substituting, we get:

y = x + 5 - [-(x - 5)] = x + 5 + x - 5 = 2x

So, y = 2x. Recall that this is the case where -5 < x < 5; multiplying all sides by 2, we get -10 < 2x = y < 10. Notice that there are 19 integers in this interval (which are -9, -8, -7, ... , 8, 9). Each one of those are possible values for y, so we get 19 possible values for y from this interval.

Finally, we have the case |x + 5| = x + 5 and |x - 5| = x - 5. Substituting:

y = x + 5 - (x - 5) = x + 5 - x + 5 = 10

So, y = 10 is yet another possibility for y.

In total, there are 1 + 19 + 1 = 21 possible values for y.
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