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

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

Case 1 : x>5

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

Case 2: -5<= x <5

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

Now x could assume 10 values in this range
-5, -4, -3, -2, -1, 0, 1, 2, 3, 4

for which y could assume 10 values
-10, -8, -6, -4, -2, 0, 2, 4, 6, 8

However since x could be a fraction , x could also assume values like:
-4.5, -3.5, -2.5, -1.5, -0.5, 0.5, 1.5, 2.5, 3.5, 4.5

for which y could take values like:
-9,-7,-5,-3,-1,1,3,5,7,9

Case 3: x< -5

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

Therefore combining all the unique values of y we see there are 21 values of y(E)

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

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

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

We see that there is a following range of values

x<-5
-5< x<5
x>5

Now for any value of x<-5 the expression will give the value -10...
Consider x=-6 so we have |-1|-|-11|= 1-11=-10
Consider x=-10, we have |-5|-|-15|= 5-15=-10
Similarly when x>5 the expression will give value of 10

Consider x=7, |12|-|2|=10
X=100,|105|-|95|=10

So we have 2 values of y possible -10, 10

Consider the range -5<x<5...At x=-5 and 5 value is -10 and 10 respectively

Consider x=-4, |-1|-|-9|=-8
x=-3,|-2|-|-8|=-6
x=-2, |-3|-|-7|=-4
Similarly x=-1 will give you different value

Like wise x=0 will give you one value and value of x from 1 to 4 you get different values of y.

So total possible values -10,10,0, + 4 values -1 to-4 and +4 values for range of x 1 to 4= 11

Ans C

This is how you solve such questions:
So just tick a box, but don't take it too serious

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?

Check Bunuel's solution here: if-y-x-5-x-5-then-y-can-take-how-many-integer-173626.html#p1378885

In the range x <= -5, you have 1 solution.
In the range -x < x < 5, you have 19 solutions.
In the range x >= 5, you have 1 solution.

Total 21 solutions.

Dear Bunuel, If the question changed to \(y=|x+5|+|x-5|\).

When \(x\leq{-5}\), then \(|x+5|=-(x+5)=-x-5\) and \(|x-5|=-(x-5)=5-x\).
Hence in this case \(y=|x+5|+|x-5|=-x-5+(5-x)=-2x\).

When \(-5<x<5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\).
Hence in this case \(y=|x+5|+|x-5|=x+5+(5-x)=10\).
1 integer values of \(y\) for this range.

When \(x\geq{5}\), then \(|x+5|=x+5\) and \(|x-5|=x-5\).
Hence in this case \(y=|x+5|+|x-5|=x+5+(x-5)=2x\).

Does the answer equal to infinite?

Yes, in this case y can take infinitely many integer values. You can see this directly from the graph: when x<-5 or when x>5, the value of y increases along with increase in absolute value of x.

There are a lot of good mathematical solutions here, but I'll be honest - even as somebody with a 790 GMAT, that's absolutely not what I'd do. I'm not confident in my ability to consistently draw correct graphs for complex absolute value functions like this one. Especially not within 2 minutes.

I would (and I did) actually just start plugging in numbers to see what happens. The biggest answer choice is 21, so there can't possibly be more than 21 values to count. That really isn't that many, if I work quickly. The numbers in the problem aren't tough either (just adding and subtracting 5). I might be able to find some patterns along the way that would speed it up as well. Plus, if I come up with an easy strategy for a problem that I know will eventually work, that means I can safely spend a little extra time on the problem. I know that I'll figure it out, so I might spend as much as 3 minutes on this one if I really had to.

Okay, let's try it:

x = 0: y = 5 - 5 = 0
x = 1: y = 6 - 4 = 2
x = 2: y = 7 - 3 = 4

Interesting - I notice that there's a pattern there. To save time, I'm going to guess that the values keep increasing until I hit x = 5, because |x + 5| will keep getting bigger, and |x - 5| will keep getting smaller. I'm going to assume that x = 3, x = 4, and x = 5 give me the values 6, 8, and 10.

If I go bigger than x = 5, what happens? How about x = 6?
|6 + 5| - |6 - 5| = 10 - interesting - that's a value I already got.
|7 + 5| - |7 - 5| = 10 - same value again. Boring. I'm going to stop testing bigger numbers.

Right now, I've got on my paper: 0, 2, 4, 6, 8, 10.

Now let's test negative numbers, since we're dealing with absolute values.

x = -1: |-1 + 5| - |-1 - 5| = 4 - 6 = -2
x = -2: |-2 + 5| - |-2 - 5| = 3 - 7 = -4

Okay, I'm going to guess that it's symmetrical, and that I can get the values -2, -4, -6, -8, and -10. If I had extra time, I could check that.

I'm almost done, but let me think just a little bit more. Is there any way I could possibly get a different value? Maybe I should try some fractions, like x = 1/2... just to see what happens.

x = 1/2: |1/2 + 5| - |1/2 - 5| = 5.5 - 4.5 = 1

Interesting! That gave me a value I didn't have before. How about x = 1/3?

x = 1/3: |1/3 + 5| - |1/3 - 5| = 5.333 - 4.6666 = not an integer

I'm thinking right now that I have to work with halves, since when I subtract, it has to come out to an integer. So x = 1/2 gives me a value of 1. x = 3/2 gives a value of 3, x = 5/2 gives a value of 5... I can probably also get -1, -3, -5, ... just by dealing with negative numbers instead.

Finally, how big can these numbers get? What about x = 9/2?

x = 9/2: |9/2 + 5| - |9/2 - 5| = 9.5 - 0.5 = 9

x = 11/2: |11/2 + 5| - |11/2 - 5| = 10.5 - 0.5 = 10

x = 13/2: |13/2 + 5| - |13/2 - 5| = 11.5-1.5 = 10

Okay, it gets 'stuck' at 10 again.

Looks like my values are:

0, 2, 4, 6, 8, 10
-2, -4, -6, -8, -10
1, 3, 5, 7, 9
-1, -3, -5, -7, -9

For a total of 21 values.

I did a lot of writing and a lot of arithmetic, but I didn't take that much time on this one. That's because I went with a plan I felt pretty confident about right from the beginning. I took the 'easy way out' - that's what you want to do on the GMAT!

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 ...

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.

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

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


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


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

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


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



\(? = 1 + 20 = 21\)



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

Regards,
Fabio.

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.

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.

The CRITICAL POINTS are where the expressions inside the absolute values are equal to 0.

x+5 = 0 when x=-5.
Substituting x=-5 into \(y=|x+5|-|x-5|\), we get:
y = |-5+5| - |-5-5| = 0-10 = -10

x-5 = 0 when x=5.
Substituting x=5 into \(y=|x+5|-|x-5|\), we get:
y = |5+5| - |5-5| = 10-0 = 10

The resulting values in blue indicate the range for y.
Thus, y can be equal to any integer value between -10 and 10, inclusive, implying that there are 21 integer options for y.


y = |x±a| - |x±b|
Given this expression, we can determine the range for y by plugging in the two critical points and solving.

Another example:
y = |x-2| - |x+10|

Here, the critical points are 2 and and -10.
Plugging x=2 into y = |x-2| - |x+10|, we get:
y = |2-2| - |2+10| = -12
Plugging x=-10 into y = |x-2| - |x+10|, we get:
y = |-10-2| - |-10+10| = 12
Thus:
-12≤y≤12