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Express 1 < x < 3 in the form |x - a| < b, where a and b are to be determined

Any help would be appreciated on how to solve this question as I am fully stuck on how to even start it.

Any help would be appreciated on how to solve this question as I am fully stuck on how to even start it.

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#2

(Original post by

Express 1 < x < 3 in the form |x - a| < b, where a and b are to be determined

Any help would be appreciated on how to solve this question as I am fully stuck on how to even start it.

**znx**)Express 1 < x < 3 in the form |x - a| < b, where a and b are to be determined

Any help would be appreciated on how to solve this question as I am fully stuck on how to even start it.

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#4

(Original post by

How would you go about finding a and b?

**znx**)How would you go about finding a and b?

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(Original post by

Simultaneous equations.

**RDKGames**)Simultaneous equations.

EDIT: I realize that would not help solve anything. Do you know of the names of these sort of questions so I can go do some revision on them?

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#6

**znx**)

Express 1 < x < 3 in the form |x - a| < b, where a and b are to be determined

Any help would be appreciated on how to solve this question as I am fully stuck on how to even start it.

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#7

(Original post by

so am I on the right lines by doing x - 3 > 1 and x - 1 < 3?

EDIT: I realize that would not help solve anything. Do you know of the names of these sort of questions so I can go do some revision on them?

**znx**)so am I on the right lines by doing x - 3 > 1 and x - 1 < 3?

EDIT: I realize that would not help solve anything. Do you know of the names of these sort of questions so I can go do some revision on them?

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(Original post by

Incredibly, I've never actually come across a question quite like this before, so I don't know what it's called. Is this in the new spec AS?

**mupsman2312**)Incredibly, I've never actually come across a question quite like this before, so I don't know what it's called. Is this in the new spec AS?

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#9

-b + a = 1

b + a = 3

therefore, b=1 and a=2.

Can you elaborate how you did that expansion please?

Edit: I just looked at it and realised I was being idiotic and forgetting the definition of what a modulus is, haha XD.

b + a = 3

therefore, b=1 and a=2.

Edit: I just looked at it and realised I was being idiotic and forgetting the definition of what a modulus is, haha XD.

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#10

(Original post by

You could try using the fact that (x-1)(x-3) = {(x-2)^2} - 1. If you apply the right steps after this, then you should end-up with the correct answer.

**mupsman2312**)You could try using the fact that (x-1)(x-3) = {(x-2)^2} - 1. If you apply the right steps after this, then you should end-up with the correct answer.

**znx**)

so am I on the right lines by doing x - 3 > 1 and x - 1 < 3?

EDIT: I realize that would not help solve anything. Do you know of the names of these sort of questions so I can go do some revision on them?

Get it?

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#11

(Original post by

This is not correct - he is not being asked to 'solve' an inequality, simply rewrite one differently.

**RDKGames**)This is not correct - he is not being asked to 'solve' an inequality, simply rewrite one differently.

Spoiler:

1 < x < 3, so (x-1)(x-3) < 0

=> x^2 - 4x + 3 < 0

=> (x^2 - 4x + 4) - 1 < 0

=> (x-2)^2 - 1 < 0

=> (x-2)^2 < 1

=> -1 < x-2 < 1, i.e. |x-2| < 1

Therefore, a = 2 and b = 1

Show

1 < x < 3, so (x-1)(x-3) < 0

=> x^2 - 4x + 3 < 0

=> (x^2 - 4x + 4) - 1 < 0

=> (x-2)^2 - 1 < 0

=> (x-2)^2 < 1

=> -1 < x-2 < 1, i.e. |x-2| < 1

Therefore, a = 2 and b = 1

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(Original post by

This is not correct - he is not being asked to 'solve' an inequality, simply rewrite one differently.

Following from my first post, adding to everything gives . This means and

Get it?

**RDKGames**)This is not correct - he is not being asked to 'solve' an inequality, simply rewrite one differently.

Following from my first post, adding to everything gives . This means and

Get it?

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#13

(Original post by

Yes that makes sense now, so is it just another type of modulus question?

**znx**)Yes that makes sense now, so is it just another type of modulus question?

There is also an alternative (and faster) way to do this question if you understand how modulus inequalities work. When you begin with your interval here is centered at . You want to center it at 0. You can subtract this value from everything and get left with which is then saying that the the difference between and 2 is between -1 and 1, which in other words says that the distance between and 2 is less than 1, hence

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(Original post by

Pretty much.

There is also an alternative (and faster) way to do this question if you understand how modulus inequalities work. When you begin with your interval here is centered at . You want to center it at 0. You can subtract this value from everything and get left with which is then saying that the the difference between and 2 is between -1 and 1, which in other words says that the distance between and 2 is less than 1, hence

**RDKGames**)Pretty much.

There is also an alternative (and faster) way to do this question if you understand how modulus inequalities work. When you begin with your interval here is centered at . You want to center it at 0. You can subtract this value from everything and get left with which is then saying that the the difference between and 2 is between -1 and 1, which in other words says that the distance between and 2 is less than 1, hence

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