andriod09
Algebra II question, a fraction on top of a fraction, please help me!
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cpattison
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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).
andriod09
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\[\huge\frac{\frac{30x^2+27x+6}{9x^2-49}}{\frac{30x+15}{30x-70}}\] That is the fraction.
andriod09
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@cpattison
andriod09
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@ganeshie8 @Hero
cpattison
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oh okay
cpattison
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one second!
Hero
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Put a divide sign between both of them, then put them next to each other. You'll know what to do then.
Hero
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\[\huge\frac{30x^2+27x+6}{9x^2-49} \div \frac{30x+15}{30x-70} \]
cpattison
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((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))
cpattison
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yeah so flip that second fraction
cpattison
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i really have to learn how to use the equation bar
andriod09
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@cpattison you should use the \\[LaTeX\] by using the \[\\] minus the \ in the middle.
cpattison
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agreed
Hero
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I just copied and pasted what @andriod09 wrote with a few minor changes :P
andriod09
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\[\huge\text i-know-how-to-use-LaTeX\]
Hero
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Not as good as me :P
andriod09
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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
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You failed to execute the proper steps. That's why you didn't get it right
Hero
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Did you flip the second fraction?
andriod09
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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
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If you had \[\frac{3}{6} \div \frac{3}{9}\] you would have no clue what to do?
andriod09
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thats simple:
\[\frac{3}{6}\div\frac{3}{9}=\frac{1.5}{1}\]
andriod09
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\[:{\]
Hero
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actually, you have it but you failed to execuate the reciprocal rule
andriod09
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\[:[\]
Hero
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\[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6} \times \frac{9}{3} = \frac{3}{2}\]
Hero
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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
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Show me the full steps of how you got 1.5 from 3/6 div 3/9 @andriod09
Hero
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You must have used a calculator bro, lol
Hero
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Because you never divide by fractions to get a decimal. Only a calculator would do that.
andriod09
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\[\frac{3}{6}\div\frac{3}{9}=\frac{3}{6}\times\frac{9}{3}=\frac{1}{2}\times3=\frac{3}{2}\]
andriod09
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btw, hero i literally typed all that out.
Hero
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Good for you. Now do your original problem in the same manner.
cpattison
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|dw:1349572930013:dw|
cpattison
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|dw:1349573068242:dw|
Hero
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Don't forget to reduce before executing
andriod09
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that doesn't help me any @cpattison
cpattison
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|dw:1349573260282:dw|
cpattison
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i'm trying to work this out too by the way
andriod09
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can you please use the \[LaTeX\]
andriod09
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just use the equation button if need be.
Hero
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Try not to do it the hard way @cpattison
cpattison
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\[\frac{ 30(10x^2+9x+2)(3x-7) }{ 15(9x^2-4)(2x+1) }\]
cpattison
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i'm not sure i know the easy way. enlighten me?
Hero
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First I'd have to make sure you're even doing this correctly
andriod09
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How is that the answer though... the answer, from the book, says that the answer is: \[\frac{2(5x+2)}{3x+7}\]
cpattison
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i don't know how that happened unless we all have done something wrong. i get \[\frac{ 2(25x^2-27x-14) }{ 9x^2-4 }\]
cpattison
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if i were you i would try and find someone in my class or around me to ask, and double check your equations
andriod09
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A)im homeschooled,
B)this is me catching up
Hero
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I was following what @cpattison did. I'm sure I would have gotten it if I had taken my own approach.
CliffSedge
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Extremes-over-means, anyone?
\[\large \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}\]
Hero
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Yep, @cpattison made a mistake because 9x^2 - 49 is in the denominator
andriod09
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cliff. dafaq/dahell does that even mean????
Hero
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|dw:1349575081973:dw|
Hero
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\[\frac{ 2(5x+2)(3x-7) }{(3x+7)(3x-7) }\]
cpattison
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oh okay. sorry for that.
Hero
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And now the 3x - 7 reduces to:
\[\frac{ 2(5x+2) }{(3x+7) }\]
CliffSedge
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@andriod09 "extremes-over-means" |dw:1349575148786:dw|
Hero
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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.
CliffSedge
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(It's the same thing . . )
andriod09
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I don't understand it.... \[:{\]
Hero
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You ignored the part where I said something "he should have already been familiar with".
CliffSedge
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Meditate on it.
CliffSedge
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I ignored nothing.
Hero
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Normal students other than @Cliffsedge learn how to change division to multiplication by flipping the second fraction.
CliffSedge
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It's the same thing . . .
CliffSedge
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|dw:1349575395748:dw|
Hero
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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.
CliffSedge
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Anyway, I was merely offering a method that might have been recognized. It wasn't, carry on.
CliffSedge
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@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
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I'm not a dog, so unfortunately I don't know how to bark.
cpattison
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none of that helps. @andriod09 is there anything else?
andriod09
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yea. i would like to understand how to work the stupid fraction problem. :{
cpattison
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did khan help at all?
Hero
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I could have sworn I just showed you. I suppose now you need help with factoring and reducing polynomials.
andriod09
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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
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@Hero are you still here??
anonymous
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\[\frac{\frac{30x^2+27x+6}{9x^2-49}}{\frac{30x+15}{30x-70}}=\frac{30x^2+27x+6}{9x^2-49}\times \frac{30x-70}{30x+15}\]
anonymous
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factor and cancel if you can
jiteshmeghwal9
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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
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for example \(9x^2-49=(3x+7)(3x-7)\)
andriod09
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ENGLISH PLEASE. K'THANKS
anonymous
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and \(30x-70=10(x-7)\) etc etc. there will be an orgy of cancellation
andriod09
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can you please type that in English for me, step by step?
andriod09
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@jiteshmeghwal9 2 things, 1) i luv your prof pic, 2) what do i do after all that> I dont get... :{
jiteshmeghwal9
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bro first factor the numerators & denominators of both fractions then tell me what do u gt ?
andriod09
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how do i do that?
jiteshmeghwal9
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what u haven't studied factorization yet
jiteshmeghwal9
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without studying factorization u can't do this question :(
andriod09
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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
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so factor \[30x^2+27x+6\]
jiteshmeghwal9
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@andriod09
andriod09
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\[30x^2+27x+6\] should i find a comon factor? because i don't think there is one. buy anyway. isn't it: \[2(5x+2)\]
jiteshmeghwal9
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\[ax^2+(a+b)x+b\]use this identity
andriod09
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thas makes no sense to me. :/
jiteshmeghwal9
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i mean break the middle term such that the sum becomes 27x & product becomes \(30x^2*6\)
andriod09
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but was the answer right? it should be right.
jiteshmeghwal9
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but after multiplying that u r not getting the right exression
jiteshmeghwal9
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\[30x^2+15x+12x+6\]\[15x(2x+1)+6(2x+1)\]\[(15x+6)(2x+1)\]
andriod09
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ohhh. okay. Thanks!
jiteshmeghwal9
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\[(3x)^2-(7)^2\]\[(3x+7)(3x-7)\]
jiteshmeghwal9
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\[30x-70\]\[10(3x-7)\]
jiteshmeghwal9
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\[30x+15\]\[15(2x+1)\]
andriod09
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TYSVM i get
it now!
andriod09
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i g2g. Bye!
jiteshmeghwal9
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\[\frac{(15x+6)(2x+1)15(2x+1)}{(3x+7)(3x-7)10(3x-7)}\]
jiteshmeghwal9
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bye & thanx for liking my profile profile pic