1. ## Simplifying Algebraic Fractions

Hello

(8x – 1) × (8x – 1)

3x________(7x + 1)

2. Originally Posted by 200001
(8x – 1) × (8x – 1)

3x________(7x + 1)
Your formatting is difficult to read. I think you mean the following:

. . . . .$\displaystyle \frac{8x\, -\, 1}{3x}\, \times\, \frac{8x\, -\, 1}{7x\, +\, 1}$

To multiply fractions, you multiply their numerators and their denominators. For instance:

. . . . .$\displaystyle \frac{1}{3}\, \times\, \frac{4}{5}\, =\, \frac{1\, \times\, 4}{3\, \times\, 5}\, =\, \frac{4}{15}$

To simplify your fraction, multiply the 8x - 1 by the 8x - 1, and the 3x by the 7x + 1.

3. Thanks

How does it wind up being:

(7x + 1)
3x

I could understand if the operation was dividing the fractions but not mulpying

If I exapnd the brackets the number becomes huge

4. Originally Posted by 200001
Thanks

How does it wind up being:

(7x + 1)
3x

I could understand if the operation was dividing the fractions but not mulpying

If I exapnd the brackets the number becomes huge
how did you get it to that? please show your work so i can help further

5. I didnt, hence why I think the question is a division over mutlipication as the answer to the question is as I have posted above.

It makes sense to invert the fraction and cancel down if its a division problem but not as a multiplication.

6. Originally Posted by 200001
I didnt, hence why I think the question is a division over mutlipication as the answer to the question is as I have posted above.

It makes sense to invert the fraction and cancel down if its a division problem but not as a multiplication.
yes, i would agree. it must be a typo.

$\displaystyle \frac{8x-1}{3x} div \frac{8x-1}{7x+1}=\frac{8x-1}{3x} * \frac{7x+1}{8x-1}=\frac{7x+1}{3x}$ is the only way it works.

7. Great
I thought I was going mad!!!