hi

the distance between 2 points is -10 NOT +10 why ...?
is this because the key points for |x+5| and for |x-5| are -5 and +5 respectively ..
please correct me if I am missing something ...

also, if we see the number line we can easily find that y=-9 as long as x= -4.5, but if we want to interact this relationship with the equation given, can it be seen as under ..?

x+5

= -4.5 + 5 = 0.5
and

x - 5

= -4.5 - 5 = -9.5

thus

|x+5| - |x-5|
= .5 - 9.5 = -9
please correct me if I am missing something...

can such problems as this one, however, be solved by recognizing that the number of integer values y can take will be the values that fall within -10<=y<=10 ....?
please correct me if I am missing something ....

anyway, please shed some light on me, giving some lessons on number lines so that I can solve with the help of number lines virtually any problem pertaining to absolute modulus ...

thanks in advance ...

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

---------------------------------------------------------------------
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
---------------------------------------------------------------------
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
---------------------------------------------------------------------

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
_________________

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.
_________________

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)

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.