Question

can someone explain how to multiply ((9x-3)/(3x+21)) / ((x+7)/(6x-2))

Answer 1

I would factor everything you can first.
(3x+21) = 3(x+7)
(9x-3) = 3(3x-1)
(6x-2) = 2(3x-1)

Answer 2

Then stuff will start canceling

Answer 3

\[\frac{\frac{9x-3}{3x+21}}{\frac{x+7}{6x-2}}*\frac{\frac{6x-2}{x+7}}{\frac{6x-2}{x+7}}\]
\[=\frac{9x-3}{3x+21} *\frac{6x-2}{x+7}\]

Answer 4

(9x-3)(6x-2)/(3x+21)(x+7)
6(3x-1)(3x-1)/3(x+7)(x+7)
2(3x-1)^2/(x+7)^2
2((3x-1)/(x+7))^2

Answer 5

\[2(3x-1)^{2}\div(x-7)^{2}\]

Answer 6

If you want a simplification, multiply top and bottom by (3x+21), then do the same with (6x-2)
\[(9x-3)/((x+7)(3x+21)/(6x-2))\]

\[(9x-3)(6x-2)/(3x ^{2}+42x+147)\]

\[(54x ^{2}-36x+6)/(3x ^{2}+42x+147)\]

You can then factorise top and bottom:

\[6(3x-1) ^{2}/3(x+7) ^{2}\]

Which gives

\[2(3x-1) ^{2}/(x+7) ^{2}\]

Answer 7

\[\large \frac{\frac{9x-3}{3x+21}}{\frac{x+7}{6x-2}} = \frac{\frac{3(3x-1)}{3(x+7)}}{\frac{x+7}{2(3x-1)}} = \frac{\frac{3x-1}{x+7}}{\frac{x+7}{2(3x-1)}} \cdot \frac{2(x+7)(3x-1)}{2(x+7)(3x-1)} \]\[\large = \frac{2(3x-1)^2}{(x+7)^2} = 2(\frac{3x-1}{x+7})^2\]

Answer 8

show off!

Answer 9

i thought you were totally against people doing all the work, polpak

Answer 10

Always when dealing with complex fractions:

To convert the complex fraction into a standard fraction multiply the top and bottom by the LCM of all the inner fractions.

Answer 11

4 other people had already done the work. I was just doing it to show an alternative method.

Answer 12

you don't delete stuff anymore?

Answer 13

And I'm more against people giving answers than I am people giving all the work. Though I think walking them through it is the best way.

Answer 14

I only delete things if an answer is given with no justification etc. And even then I cannot be everywhere so stuff gets by anyway.

Answer 15

yeah i always work it out for them if i don't walk them through it

Answer 16

peace i gotta go