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\[\huge\frac{\frac{30x^2+27x+6}{9x^2-49}}{\frac{30x+15}{30x-70}}\] That is the fraction.

oh okay

one second!

\[\huge\frac{30x^2+27x+6}{9x^2-49} \div \frac{30x+15}{30x-70} \]

yeah so flip that second fraction

i really have to learn how to use the equation bar

@cpattison you should use the \\[LaTeX\] by using the \[\\] minus the \ in the middle.

agreed

I just copied and pasted what @andriod09 wrote with a few minor changes :P

\[\huge\text i-know-how-to-use-LaTeX\]

Not as good as me :P

You failed to execute the proper steps. That's why you didn't get it right

Did you flip the second fraction?

If you had \[\frac{3}{6} \div \frac{3}{9}\] you would have no clue what to do?

thats simple:
\[\frac{3}{6}\div\frac{3}{9}=\frac{1.5}{1}\]

\[:{\]

actually, you have it but you failed to execuate the reciprocal rule

\[:[\]

\[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6} \times \frac{9}{3} = \frac{3}{2}\]

Show me the full steps of how you got 1.5 from 3/6 div 3/9 @andriod09

You must have used a calculator bro, lol

Because you never divide by fractions to get a decimal. Only a calculator would do that.

\[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6}\times\frac{9}{3}=\frac{1}{2}\times3=\frac{3}{2}\]

btw, hero i literally typed all that out.

Good for you. Now do your original problem in the same manner.

|dw:1349572930013:dw|

|dw:1349573068242:dw|

Don't forget to reduce before executing

that doesn't help me any @cpattison

|dw:1349573260282:dw|

i'm trying to work this out too by the way

can you please use the \[LaTeX\]

just use the equation button if need be.

Try not to do it the hard way @cpattison

\[\frac{ 30(10x^2+9x+2)(3x-7) }{ 15(9x^2-4)(2x+1) }\]

i'm not sure i know the easy way. enlighten me?

First I'd have to make sure you're even doing this correctly

A)im homeschooled,
B)this is me catching up

I was following what @cpattison did. I'm sure I would have gotten it if I had taken my own approach.

Extremes-over-means, anyone?
\[\large \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}\]

Yep, @cpattison made a mistake because 9x^2 - 49 is in the denominator

cliff. dafaq/dahell does that even mean????

|dw:1349575081973:dw|

\[\frac{ 2(5x+2)(3x-7) }{(3x+7)(3x-7) }\]

oh okay. sorry for that.

And now the 3x - 7 reduces to:
\[\frac{ 2(5x+2) }{(3x+7) }\]

@andriod09 "extremes-over-means" |dw:1349575148786:dw|

(It's the same thing . . )

I don't understand it.... \[:{\]

You ignored the part where I said something "he should have already been familiar with".

Meditate on it.

I ignored nothing.

It's the same thing . . .

|dw:1349575395748:dw|

Anyway, I was merely offering a method that might have been recognized. It wasn't, carry on.

I'm not a dog, so unfortunately I don't know how to bark.

none of that helps. @andriod09 is there anything else?

yea. i would like to understand how to work the stupid fraction problem. :{

did khan help at all?

factor and cancel if you can

for example \(9x^2-49=(3x+7)(3x-7)\)

ENGLISH PLEASE. K'THANKS

and \(30x-70=10(x-7)\) etc etc. there will be an orgy of cancellation

can you please type that in English for me, step by step?

@jiteshmeghwal9 2 things, 1) i luv your prof pic, 2) what do i do after all that> I dont get... :{

bro first factor the numerators & denominators of both fractions then tell me what do u gt ?

how do i do that?

what u haven't studied factorization yet

without studying factorization u can't do this question :(

so factor \[30x^2+27x+6\]

\[ax^2+(a+b)x+b\]use this identity

thas makes no sense to me. :/

i mean break the middle term such that the sum becomes 27x & product becomes \(30x^2*6\)

but was the answer right? it should be right.

but after multiplying that u r not getting the right exression

\[30x^2+15x+12x+6\]\[15x(2x+1)+6(2x+1)\]\[(15x+6)(2x+1)\]

ohhh. okay. Thanks!

\[(3x)^2-(7)^2\]\[(3x+7)(3x-7)\]

\[30x-70\]\[10(3x-7)\]

\[30x+15\]\[15(2x+1)\]

TYSVM i get
it now!

i g2g. Bye!

\[\frac{(15x+6)(2x+1)15(2x+1)}{(3x+7)(3x-7)10(3x-7)}\]

bye & thanx for liking my profile profile pic