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I have been taught that if we have a |x+b| = c. The midpoint is always -b.

I tried applying it here: so |x+5| gives us midpoint -5. and |x-5| gives us midpoint 5.

However - how do I know the actual range? I'm not actually given something like "|x+5| < 2" or whatever.

Actually, don't think of it from the mid-point perspective because then it is not useful in many circumstances. The actual logic of absolute value is this: |x - a| gives you the "distance of x from point a on the number line"

So y = |x+5|−|x−5| gives you the difference between "distance of x from -5" and "distance of x from 5" So if x = 0, difference between "distance of x from -5" and "distance of x from 5" will be "5" - "5" = 0. This gives you y = 0.

Similarly, you can handle something like this: |x+5| + |x−5| This is sum of "distance of x from -5" and "distance of x from 5".

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

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02 Sep 2016, 00:42

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?

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?

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

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12 Mar 2017, 20:12

Bunuel wrote:

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

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

When \(-5<x<5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). Hence in this case \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). Therefore for this range \(-10<(y=2x)<10\). 19 integer values of \(y\) for this range (from -9 to 9, inclusive).

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

Total = 1 + 19 + 1 = 21.

Answer: E.

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?

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

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

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

When \(-5<x<5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). Hence in this case \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). Therefore for this range \(-10<(y=2x)<10\). 19 integer values of \(y\) for this range (from -9 to 9, inclusive).

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

Total = 1 + 19 + 1 = 21.

Answer: E.

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.
_________________

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

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22 May 2017, 09:28

Bunuel wrote:

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

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

When \(-5<x<5\), then \(|x+5|=x+5\) and \(|x-5|=-(x-5)=5-x\). Hence in this case \(y=|x+5|-|x-5|=x+5-(5-x)=2x\). Therefore for this range \(-10<(y=2x)<10\). 19 integer values of \(y\) for this range (from -9 to 9, inclusive).

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

Total = 1 + 19 + 1 = 21.

Answer: E.

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

Attachment:

WolframAlpha--yx5-x-5_--2014-07-08_0728.png

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

Hi Bunuel, in many a question I have seen this graph view for abs questions. Could you please explain how to draw a graph from the abs values equations.

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.

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.

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?

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!
_________________

Chelsey Cooley | Manhattan Prep Instructor | Seattle and Online

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

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09 Jul 2017, 17:24

Can we say that in order to get the number of integer values that Y can take, we can take test cases where we will see the that maximum Y can be is 10 and minimum it can be is -10. If we test cases such as x=5, -5, and x=6 , -6 we will see the the graph will not expand nor will it converge.

Therefore the range is 20 but we need to count zero as an integer therefore, the integer values will be 21.

had it been how many integer values will x take in such a way that y will be different, then we would have 11integer values including zero.

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 mam

Thank you for your wonderful solution to the problem...

Anyway, do you believe - on the basis of number line know-how you shared here - this problem can be solved by solving only "-10<= y <=10" ..?

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 ...

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 ...

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.
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