[CT1402]

Express 2/2x-1 - 1/2x+1. - 2/4x^2 -1 as a single algebraic fraction in its lowest terms,

The anwser is 1/2x-1, but not sure how to get there.

Thanks

[notnek]

You need to find a common denominator. Using difference of two squares for the 3rd one, the three denominators are:

2x-1, 2x+1 and (2x+1)(2x-1)

What is the LCM of these three expressions?

One you've found the LCM, you need to change each fraction so that it has this LCM as its denominator.

If you're still stuck, post your working using the info. I've given you.

[steve2005]

(Original post by CT1402)
Express 2/2x-1 - 1/2x+1. - 2/4x^2 -1 as a single algebraic fraction in its lowest terms,

The anwser is 1/2x-1, but not sure how to get there.

Thanks

Why don't you use brackets?

2/(2x-1) - 1/(2x+1.) - 2/(4x^2 -1)

[CT1402]

(Original post by notnek)
You need to find a common denominator. Using difference of two squares for the 3rd one, the three denominators are:

2x-1, 2x+1 and (2x+1)(2x-1)

What is the LCM of these three expressions?

One you've found the LCM, you need to change each fraction so that it has this LCM as its denominator.

If you're still stuck, post your working using the info. I've given you.

I used the common denominator of (2x+1)(2x-1),so I got to this;

[2(2x+1)-(2x-1)-2]/[(2x+1)(2x-1)]

When I cancel it out i keep ending up with zero, not sure what I'm doing wrong?

[Imposition]

(Original post by CT1402)
Express 2/2x-1 - 1/2x+1. - 2/4x^2 -1 as a single algebraic fraction in its lowest terms,

The anwser is 1/2x-1, but not sure how to get there.

Thanks

Can you complete it?

[notnek]

(Original post by CT1402)
I used the common denominator of (2x+1)(2x-1),so I got to this;

[2(2x+1)-(2x-1)-2]/[(2x+1)(2x-1)]

When I cancel it out i keep ending up with zero, not sure what I'm doing wrong?

I've a feeling that you went straight from here to cancelling the (2x+1) and (2x-1) to get:

2-2/1

You cannot do this since (2x+1) and (2x-1) are not factors of the whole of the numerator. This is a very common GCSE error that often creeps into A Level. Please let me know if you're unsure about this and I can try to make sure you don't make the mistake again.

I'm sorry if this is not what you did but you weren't specific so I made an assumption.

From [2(2x+1)-(2x-1)-2]/[(2x+1)(2x-1)], the next step is to expand and simplify the numerator.

If you know all of this, post all of your working.

[CT1402]

(Original post by notnek)
I've a feeling that you went straight from here to cancelling the (2x+1) and (2x-1) to get:

2-2/1

You cannot do this since (2x+1) and (2x-1) are not factors of the whole of the numerator. This is a very common GCSE error that often creeps into A Level. Please let me know if you're unsure about this and I can try to make sure you don't make the mistake again.

I'm sorry if this is not what you did but you weren't specific so I made an assumption.

From [2(2x+1)-(2x-1)-2]/[(2x+1)(2x-1)], the next step is to expand and simplify
the numerator.

If you know all of this, post all of your working.

I did make that mistake, Thankyou very much I finally came to the anwser , sorry to bother you but what would the common denominator be of such;

7x-15/[2(x-2)(x-1)] - 7/2x

[notnek]

(Original post by CT1402)
I did make that mistake, Thankyou very much I finally came to the anwser , sorry to bother you but what would the common denominator be of such;

7x-15/[2(x-2)(x-1)] - 7/2x

The factors that you need to deal with are 2, x, (x-2) and (x-1). A common multiple will always be a product of these: 2x(x-2)(x-1) and since all of the factors don't share any divisors, this must be the LCM.

In short, the common denominator should be 2x(x-2)(x-1).