andriod09
  • andriod09
Algebra II question, a fraction on top of a fraction, please help me!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • andriod09
\[\huge\frac{\frac{30x^2+27x+6}{9x^2-49}}{\frac{30x+15}{30x-70}}\] That is the fraction.
andriod09
  • andriod09
@cpattison

Looking for something else?

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

More answers

andriod09
  • andriod09
@ganeshie8 @Hero
anonymous
  • anonymous
oh okay
anonymous
  • anonymous
one second!
Hero
  • Hero
Put a divide sign between both of them, then put them next to each other. You'll know what to do then.
Hero
  • Hero
\[\huge\frac{30x^2+27x+6}{9x^2-49} \div \frac{30x+15}{30x-70} \]
anonymous
  • 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
  • anonymous
yeah so flip that second fraction
anonymous
  • anonymous
i really have to learn how to use the equation bar
andriod09
  • andriod09
@cpattison you should use the \\[LaTeX\] by using the \[\\] minus the \ in the middle.
anonymous
  • anonymous
agreed
Hero
  • Hero
I just copied and pasted what @andriod09 wrote with a few minor changes :P
andriod09
  • andriod09
\[\huge\text i-know-how-to-use-LaTeX\]
Hero
  • Hero
Not as good as me :P
andriod09
  • 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
  • Hero
You failed to execute the proper steps. That's why you didn't get it right
Hero
  • Hero
Did you flip the second fraction?
andriod09
  • 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
  • Hero
If you had \[\frac{3}{6} \div \frac{3}{9}\] you would have no clue what to do?
andriod09
  • andriod09
thats simple: \[\frac{3}{6}\div\frac{3}{9}=\frac{1.5}{1}\]
andriod09
  • andriod09
\[:{\]
Hero
  • Hero
actually, you have it but you failed to execuate the reciprocal rule
andriod09
  • andriod09
\[:[\]
Hero
  • Hero
\[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6} \times \frac{9}{3} = \frac{3}{2}\]
Hero
  • 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
  • Hero
Show me the full steps of how you got 1.5 from 3/6 div 3/9 @andriod09
Hero
  • Hero
You must have used a calculator bro, lol
Hero
  • Hero
Because you never divide by fractions to get a decimal. Only a calculator would do that.
andriod09
  • andriod09
\[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6}\times\frac{9}{3}=\frac{1}{2}\times3=\frac{3}{2}\]
andriod09
  • andriod09
btw, hero i literally typed all that out.
Hero
  • Hero
Good for you. Now do your original problem in the same manner.
anonymous
  • anonymous
|dw:1349572930013:dw|
anonymous
  • anonymous
|dw:1349573068242:dw|
Hero
  • Hero
Don't forget to reduce before executing
andriod09
  • andriod09
that doesn't help me any @cpattison
anonymous
  • anonymous
|dw:1349573260282:dw|
anonymous
  • anonymous
i'm trying to work this out too by the way
andriod09
  • andriod09
can you please use the \[LaTeX\]
andriod09
  • andriod09
just use the equation button if need be.
Hero
  • Hero
Try not to do it the hard way @cpattison
anonymous
  • anonymous
\[\frac{ 30(10x^2+9x+2)(3x-7) }{ 15(9x^2-4)(2x+1) }\]
anonymous
  • anonymous
i'm not sure i know the easy way. enlighten me?
Hero
  • Hero
First I'd have to make sure you're even doing this correctly
andriod09
  • andriod09
How is that the answer though... the answer, from the book, says that the answer is: \[\frac{2(5x+2)}{3x+7}\]
anonymous
  • 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
  • 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
  • andriod09
A)im homeschooled, B)this is me catching up
Hero
  • Hero
I was following what @cpattison did. I'm sure I would have gotten it if I had taken my own approach.
anonymous
  • anonymous
Extremes-over-means, anyone? \[\large \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}\]
Hero
  • Hero
Yep, @cpattison made a mistake because 9x^2 - 49 is in the denominator
andriod09
  • andriod09
cliff. dafaq/dahell does that even mean????
Hero
  • Hero
|dw:1349575081973:dw|
Hero
  • Hero
\[\frac{ 2(5x+2)(3x-7) }{(3x+7)(3x-7) }\]
anonymous
  • anonymous
oh okay. sorry for that.
Hero
  • Hero
And now the 3x - 7 reduces to: \[\frac{ 2(5x+2) }{(3x+7) }\]
anonymous
  • anonymous
@andriod09 "extremes-over-means" |dw:1349575148786:dw|
Hero
  • 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
  • anonymous
(It's the same thing . . )
andriod09
  • andriod09
I don't understand it.... \[:{\]
Hero
  • Hero
You ignored the part where I said something "he should have already been familiar with".
anonymous
  • anonymous
Meditate on it.
anonymous
  • anonymous
I ignored nothing.
Hero
  • Hero
Normal students other than @Cliffsedge learn how to change division to multiplication by flipping the second fraction.
anonymous
  • anonymous
It's the same thing . . .
anonymous
  • anonymous
|dw:1349575395748:dw|
Hero
  • 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
  • anonymous
http://www.khanacademy.org/math/arithmetic/fractions/v/dividing-fractions this could help. he's great
anonymous
  • anonymous
Anyway, I was merely offering a method that might have been recognized. It wasn't, carry on.
anonymous
  • 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
  • Hero
I'm not a dog, so unfortunately I don't know how to bark.
anonymous
  • anonymous
none of that helps. @andriod09 is there anything else?
andriod09
  • andriod09
yea. i would like to understand how to work the stupid fraction problem. :{
anonymous
  • anonymous
did khan help at all?
Hero
  • Hero
I could have sworn I just showed you. I suppose now you need help with factoring and reducing polynomials.
andriod09
  • 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
  • andriod09
@Hero are you still here??
anonymous
  • 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
  • anonymous
factor and cancel if you can
jiteshmeghwal9
  • 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
  • anonymous
for example \(9x^2-49=(3x+7)(3x-7)\)
andriod09
  • andriod09
ENGLISH PLEASE. K'THANKS
anonymous
  • anonymous
and \(30x-70=10(x-7)\) etc etc. there will be an orgy of cancellation
andriod09
  • andriod09
can you please type that in English for me, step by step?
andriod09
  • andriod09
@jiteshmeghwal9 2 things, 1) i luv your prof pic, 2) what do i do after all that> I dont get... :{
jiteshmeghwal9
  • jiteshmeghwal9
bro first factor the numerators & denominators of both fractions then tell me what do u gt ?
andriod09
  • andriod09
how do i do that?
jiteshmeghwal9
  • jiteshmeghwal9
what u haven't studied factorization yet
jiteshmeghwal9
  • jiteshmeghwal9
without studying factorization u can't do this question :(
andriod09
  • 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
  • jiteshmeghwal9
so factor \[30x^2+27x+6\]
jiteshmeghwal9
  • jiteshmeghwal9
@andriod09
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
  • jiteshmeghwal9
\[ax^2+(a+b)x+b\]use this identity
andriod09
  • andriod09
thas makes no sense to me. :/
jiteshmeghwal9
  • jiteshmeghwal9
i mean break the middle term such that the sum becomes 27x & product becomes \(30x^2*6\)
andriod09
  • andriod09
but was the answer right? it should be right.
jiteshmeghwal9
  • jiteshmeghwal9
but after multiplying that u r not getting the right exression
jiteshmeghwal9
  • jiteshmeghwal9
\[30x^2+15x+12x+6\]\[15x(2x+1)+6(2x+1)\]\[(15x+6)(2x+1)\]
andriod09
  • andriod09
ohhh. okay. Thanks!
jiteshmeghwal9
  • jiteshmeghwal9
\[(3x)^2-(7)^2\]\[(3x+7)(3x-7)\]
jiteshmeghwal9
  • jiteshmeghwal9
\[30x-70\]\[10(3x-7)\]
jiteshmeghwal9
  • jiteshmeghwal9
\[30x+15\]\[15(2x+1)\]
andriod09
  • andriod09
TYSVM i get it now!
andriod09
  • andriod09
i g2g. Bye!
jiteshmeghwal9
  • jiteshmeghwal9
\[\frac{(15x+6)(2x+1)15(2x+1)}{(3x+7)(3x-7)10(3x-7)}\]
jiteshmeghwal9
  • jiteshmeghwal9
bye & thanx for liking my profile profile pic

Looking for something else?

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