Maths Question

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.

Note that $|x-a| < b \Rightarrow -b < x-a < b$

How would you go about finding a and b?

Simultaneous equations.

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.

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?

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?

Note that $|x-a| < b \Rightarrow -b < x-a < b$

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.

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?

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

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



Following from my first post, adding $a$ to everything gives $a-b < x < a+b$. This means $a-b=1$ and $a+b=3$

Get it?

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

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 $1<x<3$ your interval here is centered at $\frac{1+3}{2}=2$. You want to center it at 0. You can subtract this value from everything and get left with $-1 < x-2 < 1$ which is then saying that the the difference between $x$ and 2 is between -1 and 1, which in other words says that the distance between $x$ and 2 is less than 1, hence $|x-2|<1$