# Modulus

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Question: http://pmt.physicsandmathstutor.com/...dulus%20QP.pdf

Question 2

Answer: http://pmt.physicsandmathstutor.com/...dulus%20MS.pdf

My Question:

what does question 2 mean and how do you answer it?

Question 2

Answer: http://pmt.physicsandmathstutor.com/...dulus%20MS.pdf

My Question:

what does question 2 mean and how do you answer it?

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#2

Clue: What is the equivalent form of |x-a| < b. Probably want to go through your book

Edit: Your answer already tells you everything you need to know.

Edit: Your answer already tells you everything you need to know.

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(Original post by

Clue: What is the equivalent form of |x-a| < b. Probably want to go through your book

Edit: Your answer already tells you everything you need to know.

**_JB_**)Clue: What is the equivalent form of |x-a| < b. Probably want to go through your book

Edit: Your answer already tells you everything you need to know.

1 < x 2 < 1

1 < x 2 < 1

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#4

One of the inequality rules stat that: if a < b, then a - c < b - c.

Do you think you can do this? If a < b < c, then a - d < b - d < c - d.

Do you think you can do this? If a < b < c, then a - d < b - d < c - d.

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(Original post by

One of the inequality rules stat that: if a < b, then a - c < b - c.

Do you think you can do this? If a < b < c, then a - d < b - d < c - d.

**_JB_**)One of the inequality rules stat that: if a < b, then a - c < b - c.

Do you think you can do this? If a < b < c, then a - d < b - d < c - d.

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#6

(Original post by

Question: http://pmt.physicsandmathstutor.com/...dulus%20QP.pdf

Question 2

Answer: http://pmt.physicsandmathstutor.com/...dulus%20MS.pdf

My Question:

what does question 2 mean and how do you answer it?

**esmeralda123**)Question: http://pmt.physicsandmathstutor.com/...dulus%20QP.pdf

Question 2

Answer: http://pmt.physicsandmathstutor.com/...dulus%20MS.pdf

My Question:

what does question 2 mean and how do you answer it?

The value of a will just be the mid-point of the original range, as you'll want to have the same value of |x-a| for both x=1 and x=3 (namely b). Then simply substitute this value of a and x=1 or x=3 to obtain the value of b (remember that, as |x-a| > 0, b>0 also).

Alternatively, can get to the values of a and b as follows (this is probably a better way of thinking about it and setting it out):

Spoiler:

|x-a| < b is the same as saying that -b < x-a < b. You want to put the original equation into this form. You currently have 1<x<3. The width of this range is 3-1 = 2, and the width of our algebraic range is b-(-b) = b+b = 2b. The two ranges should be equivalent, so we have 2b = 2, i.e. b = 1. Thus, we now have -1 < x-a < 1. Adding a to each part of this inequality, we get a-1 < x < a+1. But, 1<x<3. Hence, a-1 = 1 and a+1 = 3; both equations yield the solution a = 2.

You can now simply double-check in your head that this solution works by substituting suitable values of x.

As you get more used to these types of questions, you should be able to see the value of a straightaway - i.e. the value of a for which substituting the limits of the original inequality will give you |(plus/minus the same value)| = (the same value). In the mark scheme, they've seen immediately that subtracting 2 from the original inequality gives you

I hope that this explanation is clear enough for you - it can be slightly tricky to explain...

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|x-a| < b is the same as saying that -b < x-a < b. You want to put the original equation into this form. You currently have 1<x<3. The width of this range is 3-1 = 2, and the width of our algebraic range is b-(-b) = b+b = 2b. The two ranges should be equivalent, so we have 2b = 2, i.e. b = 1. Thus, we now have -1 < x-a < 1. Adding a to each part of this inequality, we get a-1 < x < a+1. But, 1<x<3. Hence, a-1 = 1 and a+1 = 3; both equations yield the solution a = 2.

You can now simply double-check in your head that this solution works by substituting suitable values of x.

As you get more used to these types of questions, you should be able to see the value of a straightaway - i.e. the value of a for which substituting the limits of the original inequality will give you |(plus/minus the same value)| = (the same value). In the mark scheme, they've seen immediately that subtracting 2 from the original inequality gives you

__+__1 as the strict limits of the inequality. Hence a=2 and b=1 (as

__+__b =

__+__1).

I hope that this explanation is clear enough for you - it can be slightly tricky to explain...

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#7

So the left side is -1 and right side is +1?

|x - 2| < 1 is equivalent to -1 < x - 2 < 1.

I will use a simpler example:

|x| < 2 means the absolute value of x is less than 2. Mean x can be 1.8, -1.5 etc etc.

Then you can deduce that |x| < 2 and -2 < x < 2 is equivalent.

|x - 2| < 1 is equivalent to -1 < x - 2 < 1.

I will use a simpler example:

|x| < 2 means the absolute value of x is less than 2. Mean x can be 1.8, -1.5 etc etc.

Then you can deduce that |x| < 2 and -2 < x < 2 is equivalent.

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