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dave13
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 14 Jan 2021, 04:56
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BrentGMATPrepNow
Bunuel

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.

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
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Hi BrentGMATPrepNow :) great post just one question

why are you using only strict inequality sign, for second region i used -5 ≤ x <5

in this case -10<=x <10 can can take 20 values in second region

for third region i used non strict inequality 5 ≤ x

thank you :)

btw in what cases to use strict vs non strict inequality sign ?
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 14 Jan 2021, 12:14
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dave13

Hi BrentGMATPrepNow :) great post just one question

why are you using only strict inequality sign, for second region i used -5 ≤ x <5

in this case -10<=x <10 can can take 20 values in second region

for third region i used non strict inequality 5 ≤ x

thank you :)

btw in what cases to use strict vs non strict inequality sign ?
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Good question.
Once I tested x = 5 and x = -5 (resulting in y = 10 and y = -10), I already had two possible values of y.
So it was unnecessary to include x = 5 and x = -5 in my range of values.
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 30 Dec 2021, 22:17
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Given that \(y=|x+5|-|x-5|\) and we need to find how many integer values can y take

Now, whenever we have more than one Absolute value in the problem then we need to try to reduce the number of cases we need to solve.
We do that by putting the values inside the absolute value = 0 and then try to find the points on the number line where the value inside the absolute value will change sign

|x+5| -> x+5 = 0 => x = -5
|x-5| -> x-5 = 0 => x = 5

So, we can divide the number line into three parts

x < - 5, -5 <= x < 5, x > = 5

Attachment:
-5 to 5.JPG
-5 to 5.JPG [ 20.07 KiB | Viewed 3617 times ]

Case 1: x < - 5

=> |x+5| = -(x+5) (as x+5 will be negative) (Watch this video to learn about Basics of Absolute Value)
And |x-5| = -(x-5) (as x-5 will be negative)
=> \(y=|x+5|-|x-5|\) = -(x+5) - (-(x-5)) = -x -5 +x - 5 = -10

So, for any value of x < -5, y = -10

Case 2: -5 <= x < 5

=> |x+5| = x+5 (as x+5 will be non negative) (Watch this video to learn about Basics of Absolute Value)
And |x-5| = -(x-5) (as x-5 will be negative)
=> \(y=|x+5|-|x-5|\) = x+5 - (-(x-5)) = x + 5 + x - 5 = 2x

So, for any value of x between -5 and +5, y = 2x
So, we can take integer and decimal values with .5 in the end of x from -5 to 5 (as y=2x and 0.5*2 will give us integer)
=> x = -5, -4.5, -4, -3.5,...0, 0.5, 1..., 4.5 and corresponding values of y will be
=> y = -10, -9, -8, ....0, 1, 2, ... 9
=> 9-(-10) = 1 = 20 values

Case 3: x >=5

=> |x+5| = x+5 (as x+5 will be non negative) (Watch this video to learn about Basics of Absolute Value)
And |x-5| = x-5 (as x-5 will be non negative)
=> \(y=|x+5|-|x-5|\) = x+5 - (x-5) = x + 5 - x + 5 = 10

So, y can take 21 integer values from -10 to 10

So, Answer will be E
Hope it helps!

Watch the following video to learn How to Solve Absolute Value Problems

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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 07 Oct 2024, 08:09
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Bunuel,

Generally while solving such question, I'll consider the following ranges-
X<-5
-5<= X <5
X>= 5

I solved this question using the above mentioned ranges and the answer I got is 22.
Is is wrong for me to consider ranges in this generic manner.. I notice you've considered the first range as X<= -5 and so forth. While taking ranges do you always check the infliction point and a value on the other side of the range to decide whether you consider the = sign ?



Bunuel
SOLUTION

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

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

Considering that the transition points for \(|x+5|\) and \(|x-5|\) are -5 and 5, respectively, let's examine the function \(y=|x+5|-|x-5|\) for the following three ranges

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\). This implies that when \(x\leq{-5}\), \(y\) can take only 1 integer value, which is -10.

When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). For this range, \(-10 < (y=2x) < 10\). This implies that when \(-5 < x < 5\), \(y\) can take 19 integer values, ranging from -9 to 9, inclusive.

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\). In this case, \(y=|x+5|-|x-5|=x+5-(x-5)=10\). This implies that when \(x\geq{5}\), \(y\) can take only 1 integer value, which is 10.

Total number of integer values for \(y\) = 1 + 19 + 1 = 21.

If you are interested, here is a graph of \(y=|x+5|−|x−5|:\)



As illustrated, \(y\) is a continuous function (its values change smoothly without any gaps when moving along the domain), ranging from -10 to 10, inclusive. Consequently, \(y\) can take 21 integer values from -10 to 10, inclusive.

Try NEW absolute value DS question.

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WolframAlpha--yx5-x-5_--2014-07-08_0728.png
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 07 Oct 2024, 09:03
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shaurya_gmat
Bunuel,

Generally while solving such question, I'll consider the following ranges-
X<-5
-5<= X <5
X>= 5

I solved this question using the above mentioned ranges and the answer I got is 22.
Is is wrong for me to consider ranges in this generic manner.. I notice you've considered the first range as X<= -5 and so forth. While taking ranges do you always check the infliction point and a value on the other side of the range to decide whether you consider the = sign ?



Bunuel
SOLUTION

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

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

Considering that the transition points for \(|x+5|\) and \(|x-5|\) are -5 and 5, respectively, let's examine the function \(y=|x+5|-|x-5|\) for the following three ranges

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\). This implies that when \(x\leq{-5}\), \(y\) can take only 1 integer value, which is -10.

When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). For this range, \(-10 < (y=2x) < 10\). This implies that when \(-5 < x < 5\), \(y\) can take 19 integer values, ranging from -9 to 9, inclusive.

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\). In this case, \(y=|x+5|-|x-5|=x+5-(x-5)=10\). This implies that when \(x\geq{5}\), \(y\) can take only 1 integer value, which is 10.

Total number of integer values for \(y\) = 1 + 19 + 1 = 21.

If you are interested, here is a graph of \(y=|x+5|−|x−5|:\)



As illustrated, \(y\) is a continuous function (its values change smoothly without any gaps when moving along the domain), ranging from -10 to 10, inclusive. Consequently, \(y\) can take 21 integer values from -10 to 10, inclusive.

Try NEW absolute value DS question.

Show Spoiler
Attachment:
WolframAlpha--yx5-x-5_--2014-07-08_0728.png
Show more

When determining the intervals around the critical points, it's important to include each critical point in one of the intervals. It doesn't matter which interval includes the equal sign for a critical point, as long as all critical points are accounted for. Whether you write the intervals as x < -5, -5 ≤ x < 5, and x ≥ 5, or x ≤ -5, -5 < x ≤ 5, and x > 5, the solution will be the same. The key is to ensure every critical point is included without any gaps or overlaps.

Thus, as long as you include the equal sign at each critical point, it doesn't matter which interval it’s assigned to.
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 05 Dec 2024, 04:14
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Could you give me a X value (in the range -5<X<5) that makes Y take one of following values ? (-9,-7,-5,-3,-1,1,3,5,7,9)

Thanks

Bunuel
SOLUTION

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

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

Considering that the transition points for \(|x+5|\) and \(|x-5|\) are -5 and 5, respectively, let's examine the function \(y=|x+5|-|x-5|\) for the following three ranges

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\). This implies that when \(x\leq{-5}\), \(y\) can take only 1 integer value, which is -10.

When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). For this range, \(-10 < (y=2x) < 10\). This implies that when \(-5 < x < 5\), \(y\) can take 19 integer values, ranging from -9 to 9, inclusive.

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\). In this case, \(y=|x+5|-|x-5|=x+5-(x-5)=10\). This implies that when \(x\geq{5}\), \(y\) can take only 1 integer value, which is 10.

Total number of integer values for \(y\) = 1 + 19 + 1 = 21.

If you are interested, here is a graph of \(y=|x+5|−|x−5|:\)



As illustrated, \(y\) is a continuous function (its values change smoothly without any gaps when moving along the domain), ranging from -10 to 10, inclusive. Consequently, \(y\) can take 21 integer values from -10 to 10, inclusive.

Try NEW absolute value DS question.

Show Spoiler
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 05 Dec 2024, 04:17
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-9 = |x + 5| - |x - 5|

x = -4.5

Joaquinramirez7
Could you give me a X value (in the range -5<X<5) that makes Y take one of following values ? (-9,-7,-5,-3,-1,1,3,5,7,9)

Thanks

Bunuel
SOLUTION

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

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

Considering that the transition points for \(|x+5|\) and \(|x-5|\) are -5 and 5, respectively, let's examine the function \(y=|x+5|-|x-5|\) for the following three ranges

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\). This implies that when \(x\leq{-5}\), \(y\) can take only 1 integer value, which is -10.

When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). For this range, \(-10 < (y=2x) < 10\). This implies that when \(-5 < x < 5\), \(y\) can take 19 integer values, ranging from -9 to 9, inclusive.

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\). In this case, \(y=|x+5|-|x-5|=x+5-(x-5)=10\). This implies that when \(x\geq{5}\), \(y\) can take only 1 integer value, which is 10.

Total number of integer values for \(y\) = 1 + 19 + 1 = 21.

If you are interested, here is a graph of \(y=|x+5|−|x−5|:\)



As illustrated, \(y\) is a continuous function (its values change smoothly without any gaps when moving along the domain), ranging from -10 to 10, inclusive. Consequently, \(y\) can take 21 integer values from -10 to 10, inclusive.

Try NEW absolute value DS question.

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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 06 May 2025, 10:17
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How do we arrive at -10 < (y=2x) < 10? In the second case, -10 is still understandable since we got y=-10 when x<-5, but how did we arrive at the 10. I'm also confused as to how you came to the conclusion that there are 19 integers that can assume that value, since the equation that you derived -> -10 < (y=2x) < 10 if x is 9 or -9 or in fact even 8 or -8 we'd end up by y=2x = 9*2 = 18 > 10 not <10. Could you please break it down for me?I'm very lost. Hope I could explain my query.
Bunuel
SOLUTION

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

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

Considering that the transition points for \(|x+5|\) and \(|x-5|\) are -5 and 5, respectively, let's examine the function \(y=|x+5|-|x-5|\) for the following three ranges

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\). This implies that when \(x\leq{-5}\), \(y\) can take only 1 integer value, which is -10.

When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). For this range, \(-10 < (y=2x) < 10\). This implies that when \(-5 < x < 5\), \(y\) can take 19 integer values, ranging from -9 to 9, inclusive.

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\). In this case, \(y=|x+5|-|x-5|=x+5-(x-5)=10\). This implies that when \(x\geq{5}\), \(y\) can take only 1 integer value, which is 10.

Total number of integer values for \(y\) = 1 + 19 + 1 = 21.

If you are interested, here is a graph of \(y=|x+5|−|x−5|:\)



As illustrated, \(y\) is a continuous function (its values change smoothly without any gaps when moving along the domain), ranging from -10 to 10, inclusive. Consequently, \(y\) can take 21 integer values from -10 to 10, inclusive.

Try NEW absolute value DS question.

Show Spoiler
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 06 May 2025, 10:30
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Chahel
How do we arrive at -10 < (y=2x) < 10? In the second case, -10 is still understandable since we got y=-10 when x<-5, but how did we arrive at the 10. I'm also confused as to how you came to the conclusion that there are 19 integers that can assume that value, since the equation that you derived -> -10 < (y=2x) < 10 if x is 9 or -9 or in fact even 8 or -8 we'd end up by y=2x = 9*2 = 18 > 10 not <10. Could you please break it down for me?I'm very lost. Hope I could explain my query.
Bunuel
SOLUTION

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

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

Considering that the transition points for \(|x+5|\) and \(|x-5|\) are -5 and 5, respectively, let's examine the function \(y=|x+5|-|x-5|\) for the following three ranges

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\). This implies that when \(x\leq{-5}\), \(y\) can take only 1 integer value, which is -10.

When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). For this range, \(-10 < (y=2x) < 10\). This implies that when \(-5 < x < 5\), \(y\) can take 19 integer values, ranging from -9 to 9, inclusive.

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\). In this case, \(y=|x+5|-|x-5|=x+5-(x-5)=10\). This implies that when \(x\geq{5}\), \(y\) can take only 1 integer value, which is 10.

Total number of integer values for \(y\) = 1 + 19 + 1 = 21.

If you are interested, here is a graph of \(y=|x+5|−|x−5|:\)



As illustrated, \(y\) is a continuous function (its values change smoothly without any gaps when moving along the domain), ranging from -10 to 10, inclusive. Consequently, \(y\) can take 21 integer values from -10 to 10, inclusive.

Try NEW absolute value DS question.

Show Spoiler
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1. Why is it -10 < y = 2x < 10?
Because in the second case we have -5 < x < 5, so multiplying both sides by 2 gives -10 < (2x = y) < 10.

2. Why are there 19 integer values?
These are the integers strictly between -10 and 10, which means from -9 to 9 inclusive, making 19 values.

Please review the full discussion for better understanding. The solution is elaborated in several posts there.
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 26 May 2025, 13:20
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Why does squaring everything not work? Thnx
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Bunuel
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 26 May 2025, 13:28
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eps018us
y = |x + 5| - |x - 5|

Why does squaring everything not work? Thnx
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Squaring is usually done to eliminate absolute values and simplify the inequality. But here, after squaring, the absolute values are still there, so squaring doesn't really help. Also, when squared, we get y^2 on the left-hand side, while we are asked to find the value that y can take, so this further complicates the matter. So, better to study several different approaches that work, as shown in the thread.
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 21 Jun 2025, 09:07
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what if we square it on both sides, and take = y^2 = a^2 - b^2, a = x+5, b = x-5. Can we solve it like that? if not, then why not?
Bunuel
If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?

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

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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 21 Jun 2025, 10:03
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Dipanshi_25
what if we square it on both sides, and take = y^2 = a^2 - b^2, a = x+5, b = x-5. Can we solve it like that? if not, then why not?
Bunuel
If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?

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

M30-21

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Squaring will not eliminate the square root sign, you'd get:

y^2 = 2x^2 - 2|x + 5| * |x - 5| + 50

So, no.
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 01 Dec 2025, 09:06
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Bunuel, would you say that this is too difficult for GMAT level and should be tagged as such? Or is thsi difficulty possible on GMAT?

Bunuel
SOLUTION

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

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

Considering that the transition points for \(|x+5|\) and \(|x-5|\) are -5 and 5, respectively, let's examine the function \(y=|x+5|-|x-5|\) for the following three ranges

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=-x-5-(5-x)=-10\). This implies that when \(x\leq{-5}\), \(y\) can take only 1 integer value, which is -10.

When \(-5 < x < 5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). In this case, \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). For this range, \(-10 < (y=2x) < 10\). This implies that when \(-5 < x < 5\), \(y\) can take 19 integer values, ranging from -9 to 9, inclusive.

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\). In this case, \(y=|x+5|-|x-5|=x+5-(x-5)=10\). This implies that when \(x\geq{5}\), \(y\) can take only 1 integer value, which is 10.

Total number of integer values for \(y\) = 1 + 19 + 1 = 21.

If you are interested, here is a graph of \(y=|x+5|−|x−5|:\)



As illustrated, \(y\) is a continuous function (its values change smoothly without any gaps when moving along the domain), ranging from -10 to 10, inclusive. Consequently, \(y\) can take 21 integer values from -10 to 10, inclusive.

Try NEW absolute value DS question.

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WolframAlpha--yx5-x-5_--2014-07-08_0728.png
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 01 Dec 2025, 10:59
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KyleKouton
Bunuel, would you say that this is too difficult for GMAT level and should be tagged as such? Or is thsi difficulty possible on GMAT?


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As the stats in the original post show, and as the difficulty tag indicates, this is an 805+ level question, so it’s very hard. But the underlying concepts are fully aligned with what GMAT tests, and something at this could appear on the exam.
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 24 Mar 2026, 02:36
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This question got me too. It asks integer value of Y and not X. for values of X such as 4.5, 3.5 ..... -3.5, -4.5 , Y will still be integer :)

La1yaMalhotra
In the range -5<x<=5, there are only 11 integers. So, if we are solving for x using -10<=2x<10 and it gives us 20 integers,then only 11 of these are possible within the range -5<x<=5. Can someone please help me understand as to why we are going with 21 as the answer?
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If y = |x + 5| - |x - 5|, then y can take how many integer vales? 24 Mar 2026, 06:57
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If \(y=|x+5|-|x-5|\), then \(y\) can take how many integer values?

y=|x+5|-|x-5|

Case 1: x <= -5
y = -x-5 - (5-x) = -x-5-5+x = -10

Case 2: -5 < x <= 5
y = x+5 - (5-x) = x+5-5+x = 2x
y = {-8,-6,-4,-2,0,2,4,6,8,10}

Case 3: x>5
y = x+5-(x-5) = x+5-x+5 = 10

Combining all above
y = {-10,-8,-6,-4,-2,0,2,4,6,8,10} : 21 values

IMO E
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