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# If y = |x + 5| - |x - 5|, then y can take how many integer vales?

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Re: If y = |x + 5| - |x - 5|, then y can take how many integer vales?  [#permalink]

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30 Jul 2017, 05:24
VeritasPrepKarishma wrote:
Bunuel wrote:

If $$y=|x+5|-|x-5|$$, then $$y$$ can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21

Kudos for a correct solution.

Use the number line for mod questions:

___________-5 ______________0______________5___________

You want the values of y which is the difference between "distance from -5" and "distance from 5".

Anywhere on the left of -5,

<------------------------------------------------
<-------------
___________-5 ______________0______________5___________

The difference between the two distances will always be -10 (taking difference to mean what it does for GMAT)

In between -5 and 5, can y can the value of -9? Sure. Move 0.5 to the right of -5. At x = -4.5, distance from -5 will be .5 and distance from 5 will be 9.5. So y = 0.5 - 9.5 = -9
Note that x needn't be an integer. Only y needs to be an integer.

Similarly, at various points between -5 and 5, y will take all integer values from -9 to 9.

To the right of 5, the difference between the two distances will always be 10.

So values that y can take: -10, -9, ... 0 ... 9, 10 i.e. a total of 21 values.

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 ...

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Re: If y = |x + 5| - |x - 5|, then y can take how many integer vales?  [#permalink]

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31 Jul 2017, 23:52
ssislam wrote:
VeritasPrepKarishma wrote:
Bunuel wrote:

If $$y=|x+5|-|x-5|$$, then $$y$$ can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21

Kudos for a correct solution.

Use the number line for mod questions:

___________-5 ______________0______________5___________

You want the values of y which is the difference between "distance from -5" and "distance from 5".

Anywhere on the left of -5,

<------------------------------------------------
<-------------
___________-5 ______________0______________5___________

The difference between the two distances will always be -10 (taking difference to mean what it does for GMAT)

In between -5 and 5, can y can the value of -9? Sure. Move 0.5 to the right of -5. At x = -4.5, distance from -5 will be .5 and distance from 5 will be 9.5. So y = 0.5 - 9.5 = -9
Note that x needn't be an integer. Only y needs to be an integer.

Similarly, at various points between -5 and 5, y will take all integer values from -9 to 9.

To the right of 5, the difference between the two distances will always be 10.

So values that y can take: -10, -9, ... 0 ... 9, 10 i.e. a total of 21 values.

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 ...

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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Re: If y = |x + 5| - |x - 5|, then y can take how many integer vales?  [#permalink]

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30 Sep 2018, 07:51
Top Contributor
1
Bunuel wrote:
If $$y=|x+5|-|x-5|$$, then $$y$$ can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21

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.

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

Cheers,
Brent
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Re: If y = |x + 5| - |x - 5|, then y can take how many integer vales?  [#permalink]

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19 Dec 2018, 08:30
Bunuel wrote:

If $$y=|x+5|-|x-5|$$, then $$y$$ can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21

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

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

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

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

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

$$? = 1 + 20 = 21$$

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

Regards,
Fabio.
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Re: If y = |x + 5| - |x - 5|, then y can take how many integer vales?  [#permalink]

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19 Dec 2018, 09:36
Bunuel wrote:

If anyone interested here is a graph of $$y=|x+5|-|x-5|$$:

As you can see y is a continuous function from -10 to 10, inclusive.

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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Re: If y = |x + 5| - |x - 5|, then y can take how many integer vales?  [#permalink]

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31 Aug 2019, 04:26
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.
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Re: If y = |x + 5| - |x - 5|, then y can take how many integer vales?  [#permalink]

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11 Sep 2019, 19:15
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)
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Re: If y = |x + 5| - |x - 5|, then y can take how many integer vales?  [#permalink]

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22 Sep 2019, 07:27
Bunuel wrote:

If $$y=|x+5|-|x-5|$$, then $$y$$ can take how many integer values?

A. 5
B. 10
C. 11
D. 20
E. 21

Kudos for a correct solution.

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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Re: If y = |x + 5| - |x - 5|, then y can take how many integer vales?  [#permalink]

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08 Jan 2020, 00:03
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.
Re: If y = |x + 5| - |x - 5|, then y can take how many integer vales?   [#permalink] 08 Jan 2020, 00:03

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