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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.

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

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

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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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.

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

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

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

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

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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Why does squaring everything not work? Thnx
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y = |x + 5| - |x - 5|

Why does squaring everything not work? Thnx

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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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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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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