Solve this fraction: 3x^2/9x-9 x 4x-4/2x

[Furai]

Hi guys, i've been stuck on this for quite a while. It is quite an advanced fraction question and it's got me stumped

3x^2/9x-9 x 4x-4/2x

I have to try and write this as one fraction in its most basic (simple) form

Am not sure how am supposed to do this though as my information on manipulating fractions is quite hazy :/

Any help would be much appreciated!

Thanks

 

[ksdhart2]

As written, your problem looks like this:

\(\displaystyle \dfrac{3x^2}{9x}-9 \cdot 4x-\dfrac{4}{2x}\)

Which should be a simple enough matter of cancelling like terms in the numerator and denominator. However, I suspect that what you meant was this:

\(\displaystyle \dfrac{3x^2}{9x-9} \cdot \dfrac{4x-4}{2x}\)

To write this as text, you'd write it as: (3x^2)/(9x-9) * (4x-4)/(2x). Make sure you understand why these grouping symbols are important.

As for how to solve this problem, my first step would be to multiply the fractions together to get a single fraction. What do you get when you do that? If you need a refresher on the basic rules of fraction multiplication, try here.

 

[Furai]

I multiplied the fractions and ended up with:

(12x^3 - 12x^2) / (18x^2 - 18x)


How would i simplfy?

 

[Otis]

3x^2/9x-9 x 4x-4/2x

Please type an asterisk to show multiplication, especially when the letter x is being used for something else.

3x^2/(9x-9) * (4x-4)/(2x)

Cheers

 

[ksdhart2]

I multiplied the fractions and ended up with:

(12x^3 - 12x^2) / (18x^2 - 18x)


How would i simplfy?

Try factoring both the numerator and the denominator. Then are there any common terms you can cancel? What's left over after you do that?

 

[Otis]

1st step: cancel an "x":
3x / (9x - 9) * (4x - 4) / 2

2nd step: factor:
3x / (9(x - 1)) * 4(x - 1) / 2

:idea: If you factor first, then you can also cancel 2s and 3s.