Algebra II question, a fraction on top of a fraction, please help me!

- andriod09

Algebra II question, a fraction on top of a fraction, please help me!

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

a fraction on top of a fraction can be rewritten as say (2/3)/(3/4) then you can flip the second one because to divide you multiply by the reciprocal (2/3)*(4/3) then you multiply across the fractions (2*4)/(3*3).

- andriod09

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

- andriod09

@cpattison

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- andriod09

@ganeshie8 @Hero

- anonymous

oh okay

- anonymous

one second!

- Hero

Put a divide sign between both of them, then put them next to each other. You'll know what to do then.

- Hero

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

- anonymous

((30x2+27x+6)/(9x2−49))/((30x+15)/(30x−70)) is where we start.
then you can flip the bottom half and multiply ((30x2+27x+6)/(9x2−49))*((30x−70)/(30x+15)) then you just multiply though ((30x2+27x+6)*(30x−70))/((9x2−49)*(30x+15))

- anonymous

yeah so flip that second fraction

- anonymous

i really have to learn how to use the equation bar

- andriod09

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

- anonymous

agreed

- Hero

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

- andriod09

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

- Hero

Not as good as me :P

- andriod09

the answer isn't supposed to be like that. it is supposed to be on top of each other though. thats what it says. the answer it:
\[\frac{2(5x+2)}{(x-7)}\]

- Hero

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

- Hero

Did you flip the second fraction?

- andriod09

no. i don't know how to solve it like that, you have to factor it out is what the book says, but i don't get it.

- Hero

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

- andriod09

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

- andriod09

\[:{\]

- Hero

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

- andriod09

\[:[\]

- Hero

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

- Hero

But really, you're supposed to do the reciprocal then reduce in this manner:
\[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6} \times \frac{9}{3} =\frac{1}{2} \times 3= \frac{3}{2}\]

- Hero

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

- Hero

You must have used a calculator bro, lol

- Hero

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

- andriod09

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

- andriod09

btw, hero i literally typed all that out.

- Hero

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

- anonymous

|dw:1349572930013:dw|

- anonymous

|dw:1349573068242:dw|

- Hero

Don't forget to reduce before executing

- andriod09

that doesn't help me any @cpattison

- anonymous

|dw:1349573260282:dw|

- anonymous

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

- andriod09

can you please use the \[LaTeX\]

- andriod09

just use the equation button if need be.

- Hero

Try not to do it the hard way @cpattison

- anonymous

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

- anonymous

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

- Hero

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

- andriod09

How is that the answer though... the answer, from the book, says that the answer is: \[\frac{2(5x+2)}{3x+7}\]

- anonymous

i don't know how that happened unless we all have done something wrong. i get \[\frac{ 2(25x^2-27x-14) }{ 9x^2-4 }\]

- anonymous

if i were you i would try and find someone in my class or around me to ask, and double check your equations

- andriod09

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

- Hero

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

- anonymous

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

- Hero

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

- andriod09

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

- Hero

|dw:1349575081973:dw|

- Hero

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

- anonymous

oh okay. sorry for that.

- Hero

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

- anonymous

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

- Hero

I already know about means over extremes. I was just trying to explain it in a way that @andriod09 should have already been familiar with.

- anonymous

(It's the same thing . . )

- andriod09

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

- Hero

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

- anonymous

Meditate on it.

- anonymous

I ignored nothing.

- Hero

Normal students other than @Cliffsedge learn how to change division to multiplication by flipping the second fraction.

- anonymous

It's the same thing . . .

- anonymous

|dw:1349575395748:dw|

- Hero

I know that bro, but you should have registered that I was trying to show the poor kid something he should have already been familiar with. Ask the majority of students at a school what means over extremes mean and they'll give you a blank stare.

- anonymous

http://www.khanacademy.org/math/arithmetic/fractions/v/dividing-fractions this could help. he's great

- anonymous

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

- anonymous

@Hero , but did you notice that I didn't just pop in and say, "extremes-over-means, yo!" and leave? I gave a formula that should be immediately recognized by the more modern phrasing of "keep-flip-change."
I'll just leave now, because I don't want you to continue derailing the thread just to bark at me.

- Hero

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

- anonymous

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

- andriod09

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

- anonymous

did khan help at all?

- Hero

I could have sworn I just showed you. I suppose now you need help with factoring and reducing polynomials.

- andriod09

cp, i don't havethe time for khan, but yes, Sam is a really good tutor. i have been to that site nealy a year now. And Hero, YES PLEASE.

- andriod09

@Hero are you still here??

- anonymous

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

- anonymous

factor and cancel if you can

- jiteshmeghwal9

A type of complex fraction\[\huge\frac{\frac{30x^2+27x+6}{9x^2-49}}{\frac{30x+15}{30x-70}}\]\[\LARGE{\frac{30x^2+27x+6}{9x^2-49} \div \frac{30x+15}{30x-70}}\]\[\LARGE{\frac{30x^2+27x+6}{9x^2-49} \times \frac{30x-70}{30x+15}}\]

- anonymous

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

- andriod09

ENGLISH PLEASE. K'THANKS

- anonymous

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

- andriod09

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

- andriod09

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

- jiteshmeghwal9

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

- andriod09

how do i do that?

- jiteshmeghwal9

what u haven't studied factorization yet

- jiteshmeghwal9

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

- andriod09

not for fractions. i can factor \[x^2+10x+20\] easily. just not for fractions. plus, im homeschooled, so things aren't the same for me.

- jiteshmeghwal9

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

- jiteshmeghwal9

@andriod09

- andriod09

\[30x^2+27x+6\] should i find a comon factor? because i don't think there is one. buy anyway. isn't it: \[2(5x+2)\]

- jiteshmeghwal9

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

- andriod09

thas makes no sense to me. :/

- jiteshmeghwal9

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

- andriod09

but was the answer right? it should be right.

- jiteshmeghwal9

but after multiplying that u r not getting the right exression

- jiteshmeghwal9

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

- andriod09

ohhh. okay. Thanks!

- jiteshmeghwal9

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

- jiteshmeghwal9

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

- jiteshmeghwal9

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

- andriod09

TYSVM i get
it now!

- andriod09

i g2g. Bye!

- jiteshmeghwal9

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

- jiteshmeghwal9

bye & thanx for liking my profile profile pic

Looking for something else?

Not the answer you are looking for? Search for more explanations.