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

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

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oh okay
one second!
Put a divide sign between both of them, then put them next to each other. You'll know what to do then.
\[\huge\frac{30x^2+27x+6}{9x^2-49} \div \frac{30x+15}{30x-70} \]
((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))
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.
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
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: \[\frac{2(5x+2)}{(x-7)}\]
You failed to execute the proper steps. That's why you didn't get it right
Did you flip the second fraction?
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.
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
\[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6} \times \frac{9}{3} = \frac{3}{2}\]
But really, you're supposed to do the reciprocal then reduce in this manner: \[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6} \times \frac{9}{3} =\frac{1}{2} \times 3= \frac{3}{2}\]
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.
btw, hero i literally typed all that out.
Good for you. Now do your original problem in the same manner.
Don't forget to reduce before executing
that doesn't help me any @cpattison
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
How is that the answer though... the answer, from the book, says that the answer is: \[\frac{2(5x+2)}{3x+7}\]
i don't know how that happened unless we all have done something wrong. i get \[\frac{ 2(25x^2-27x-14) }{ 9x^2-4 }\]
if i were you i would try and find someone in my class or around me to ask, and double check your equations
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????
\[\frac{ 2(5x+2)(3x-7) }{(3x+7)(3x-7) }\]
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|
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.
(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.
Normal students other than @Cliffsedge learn how to change division to multiplication by flipping the second fraction.
It's the same thing . . .
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. this could help. he's great
Anyway, I was merely offering a method that might have been recognized. It wasn't, carry on.
@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.
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?
I could have sworn I just showed you. I suppose now you need help with factoring and reducing polynomials.
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.
@Hero are you still here??
\[\frac{\frac{30x^2+27x+6}{9x^2-49}}{\frac{30x+15}{30x-70}}=\frac{30x^2+27x+6}{9x^2-49}\times \frac{30x-70}{30x+15}\]
factor and cancel if you can
A type of complex fraction\[\huge\frac{\frac{30x^2+27x+6}{9x^2-49}}{\frac{30x+15}{30x-70}}\]\[\LARGE{\frac{30x^2+27x+6}{9x^2-49} \div \frac{30x+15}{30x-70}}\]\[\LARGE{\frac{30x^2+27x+6}{9x^2-49} \times \frac{30x-70}{30x+15}}\]
for example \(9x^2-49=(3x+7)(3x-7)\)
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 :(
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.
so factor \[30x^2+27x+6\]
\[30x^2+27x+6\] should i find a comon factor? because i don't think there is one. buy anyway. isn't it: \[2(5x+2)\]
\[ax^2+(a+b)x+b\]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
ohhh. okay. Thanks!
TYSVM i get it now!
i g2g. Bye!
bye & thanx for liking my profile profile pic

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