## andriod09 3 years ago Algebra II question, a fraction on top of a fraction, please help me!

1. cpattison

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).

2. andriod09

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

3. andriod09

@cpattison

4. andriod09

@ganeshie8 @Hero

5. cpattison

oh okay

6. cpattison

one second!

7. Hero

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

8. Hero

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

9. cpattison

((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))

10. cpattison

yeah so flip that second fraction

11. cpattison

i really have to learn how to use the equation bar

12. andriod09

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

13. cpattison

agreed

14. Hero

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

15. andriod09

$\huge\text i-know-how-to-use-LaTeX$

16. Hero

Not as good as me :P

17. 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: $\frac{2(5x+2)}{(x-7)}$

18. Hero

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

19. Hero

Did you flip the second fraction?

20. 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.

21. Hero

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

22. andriod09

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

23. andriod09

$:{$

24. Hero

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

25. andriod09

$:[$

26. Hero

$\frac{3}{6}\div\frac{3}{9}=\frac{3}{6} \times \frac{9}{3} = \frac{3}{2}$

27. Hero

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

28. Hero

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

29. Hero

You must have used a calculator bro, lol

30. Hero

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

31. andriod09

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

32. andriod09

btw, hero i literally typed all that out.

33. Hero

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

34. cpattison

|dw:1349572930013:dw|

35. cpattison

|dw:1349573068242:dw|

36. Hero

Don't forget to reduce before executing

37. andriod09

that doesn't help me any @cpattison

38. cpattison

|dw:1349573260282:dw|

39. cpattison

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

40. andriod09

can you please use the $LaTeX$

41. andriod09

just use the equation button if need be.

42. Hero

Try not to do it the hard way @cpattison

43. cpattison

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

44. cpattison

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

45. Hero

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

46. andriod09

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

47. cpattison

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

48. cpattison

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

49. andriod09

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

50. Hero

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

51. CliffSedge

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

52. Hero

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

53. andriod09

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

54. Hero

|dw:1349575081973:dw|

55. Hero

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

56. cpattison

oh okay. sorry for that.

57. Hero

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

58. CliffSedge

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

59. 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.

60. CliffSedge

(It's the same thing . . )

61. andriod09

I don't understand it.... $:{$

62. Hero

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

63. CliffSedge

Meditate on it.

64. CliffSedge

I ignored nothing.

65. Hero

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

66. CliffSedge

It's the same thing . . .

67. CliffSedge

|dw:1349575395748:dw|

68. 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.

69. cpattison

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

70. CliffSedge

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

71. CliffSedge

@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.

72. Hero

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

73. cpattison

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

74. andriod09

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

75. cpattison

did khan help at all?

76. Hero

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

77. 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.

78. andriod09

@Hero are you still here??

79. satellite73

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

80. satellite73

factor and cancel if you can

81. 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}}$

82. satellite73

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

83. andriod09

84. satellite73

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

85. andriod09

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

86. andriod09

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

87. jiteshmeghwal9

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

88. andriod09

how do i do that?

89. jiteshmeghwal9

what u haven't studied factorization yet

90. jiteshmeghwal9

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

91. 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.

92. jiteshmeghwal9

so factor $30x^2+27x+6$

93. jiteshmeghwal9

@andriod09

94. andriod09

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

95. jiteshmeghwal9

$ax^2+(a+b)x+b$use this identity

96. andriod09

thas makes no sense to me. :/

97. jiteshmeghwal9

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

98. andriod09

but was the answer right? it should be right.

99. jiteshmeghwal9

but after multiplying that u r not getting the right exression

100. jiteshmeghwal9

$30x^2+15x+12x+6$$15x(2x+1)+6(2x+1)$$(15x+6)(2x+1)$

101. andriod09

ohhh. okay. Thanks!

102. jiteshmeghwal9

$(3x)^2-(7)^2$$(3x+7)(3x-7)$

103. jiteshmeghwal9

$30x-70$$10(3x-7)$

104. jiteshmeghwal9

$30x+15$$15(2x+1)$

105. andriod09

TYSVM i get it now!

106. andriod09

i g2g. Bye!

107. jiteshmeghwal9

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

108. jiteshmeghwal9

bye & thanx for liking my profile profile pic