## anonymous one year ago find the LCD of booth fractions and then simplify the expression completly assume the denominator equils 0 2x 3y ____ - ______ 6x^3y 4xy^2

1. anonymous

this could help with finding the LCD http://www.helpwithfractions.com/math-homework-helper/least-common-denominator/

2. anonymous

ok let me check it out

3. anonymous

can you do live help please

4. anonymous

live help?

5. anonymous

yea like help me now walk through it with me it is easyer

6. anonymous

okay. well for this question, when you say to assume that the denominator equals zero, does that mean to just completely avoid the bottom of each fractions?

7. anonymous

idk

8. Michele_Laino

we can rewrite your expression as below: $\Large \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{1}{{3{x^2}y}} - \frac{3}{{4xy}}$

9. anonymous

what did u take out?

10. anonymous

The LCD is the smallest number or expression that is divisible by both denominators. You need to take each part of it at a time. Let's look at the numbers, then the x's then the y's. What is the smallest number that is divisible by both 3 and 6. In other words, what is the LCM of 3 and 6?

11. anonymous

3

12. anonymous

No, it is 6

13. Michele_Laino

we have to compute the LCM between: $\Large 3{x^2}y,\quad and\quad 4xy$

14. anonymous

This means that the number part we need for the LCD is 6. Now let's look at the x's. There is an x^2 and an x. The lowest expression that will divide both x^2 and x is x^2, so we need x^2 for the LCD. Now let's look at the y part. You have y and y^3. The simplest expression with y that will divide evenly by y and y^3 is y^3, so you need y^3 in the LCD. Once you put together the number part, the x part and rthe y part, you end up with: LCD = 6x^2y^3

15. anonymous

I am so confused I have 2 diffrent ppl telling me 2 different things

16. anonymous

I'm not sure where the other person is going with their answer but mine are clear and simple

17. anonymous

ok

18. Michele_Laino

I'm sure that my procedure is the right procedure

19. anonymous

.... aren't you a qualified helper lol

20. anonymous

ook

21. anonymous

but mine is correct I'm sure

22. Michele_Laino

23. anonymous

I want to know what each of ur answers r

24. anonymous

@Michele_Laino thank you

25. anonymous

26. anonymous

ok

27. anonymous

wait I'm sorry, I used the wrong number

28. anonymous

ok

29. anonymous

this is your equation right? $\frac{ 2 }{ 6x ^{3y} } -\frac{ 3y }{ 4xy ^{2}}$

30. anonymous

okay so what is the LCD of 6 and 4? figure that out first

31. anonymous

3*2^2

32. anonymous

?

33. anonymous

the LCD

34. anonymous

so 12

35. anonymous

Correct. The smallest number that is divisible by both 4 and 6 is 12, That means the number part we need for the LCD is 12

36. anonymous

then what?

37. anonymous

Now let's look at the x's. There is an x^3 and an x. The lowest expression that will divide both x^3 and x is x^3, so we need x^3 for the LCD.

38. anonymous

Now let's look at the y part. You have y and y^2. The simplest expression with y that will divide evenly by y and y^2 is y^2, so you need y^2 in the LCD.

39. anonymous

ok so it is 12x^3y^2

40. anonymous

yes!!

41. Michele_Laino

that's right!

42. anonymous

:) thank you

43. Michele_Laino

good job! @aaldia678

44. anonymous

hb the numerators?

45. anonymous

12x^3y^2 is the final answer right. you just needed the LCD

46. anonymous

whatttt

47. Michele_Laino

we have to simplify the expression above

48. anonymous

btw it is assume no denomiator =0

49. Michele_Laino

yes!

50. anonymous

so what is the numerator if we found the denominator

51. anonymous

$\left(\begin{matrix}2x \\ 0\end{matrix}\right) - \frac{ 3y }{ 0}$ so this is the equation that needs to be simplified?

52. anonymous

sorry, that first part came out weird

53. anonymous

no denominator =0 and thats ok

54. Michele_Laino

no one of the denominators can be equal to zero @aaldia678

55. anonymous

thx @Michele_Laino

56. anonymous

hmm okay so what would be the full answer?

57. Michele_Laino

step #1 we have to compute this division: $\Large \frac{{12{x^3}{y^2}}}{{2x}} = ...?$

58. anonymous

my question is why is the 12 term the numerator?

59. anonymous

that equals 6x^2y^2

60. Michele_Laino

ok! now we have to multiply the first numerator by 6x^2y^2, namely, what is: $\Large 2x \cdot 6{x^2}{y^2} = ...?$

61. anonymous

12x^3y^2

62. Michele_Laino

here is wha we are doing: |dw:1432936812802:dw|

63. anonymous

is that a 72?

64. Michele_Laino

please wait a moment the least common multiple is: $\Large 24{x^3}{y^2}$

65. anonymous

ok

66. Michele_Laino

not 12x^3y^2

67. anonymous

ohhh ok

68. Michele_Laino

ok! let's restart

69. anonymous

ok clean slate

70. Michele_Laino

we have to compute this: $\Large \frac{{24{x^3}{y^2}}}{{2x}} = ...?$

71. anonymous

can we start from the very beginging?

72. Michele_Laino

sorry I have made an error, we have to compute this quantity: $\frac{{24{x^3}{y^2}}}{{6{x^3}y}} = ...?$

73. Michele_Laino

yes! we are start from the beginning

74. Michele_Laino

starting*

75. anonymous

ok :) how did u get the 24?

76. Michele_Laino

since we have to factorize 6 and 4, so we get: 6 = 2*3 4=2^2*3 now we have to pick all common and uncommon factor with the highest exponent, so we pick 2^2 and 3, then we have: 2^2*3=12 so sorry @aaldia678 is right! the least common multiple is: 12 x^3 y^2

77. Michele_Laino

next step: $\frac{{12{x^3}{y^2}}}{{6{x^3}y}} = 2y$

78. anonymous

yeah lol I was right

79. anonymous

so do you guys know the final answer yet?

80. Michele_Laino

then we have to multiply the first numerator by 2y, so we can write: $\Large \frac{{2x \cdot 2y}}{{12{x^3}{y^2}}}$ that is the first term

81. Michele_Laino

then, next we have to compute this: $\Large \frac{{12{x^3}{y^2}}}{{4x{y^2}}} = 3{x^2}$

82. anonymous

ok

83. Michele_Laino

then we have to multiply the second numerator by 3x^2, so we can write: $\Large \frac{{3{x^2} \cdot 3y}}{{12{x^3}{y^2}}}$ that is the second term

84. Michele_Laino

so we got this expression: $\Large \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right) - \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}}$

85. anonymous

ok

86. Michele_Laino

then we have to simplify, and after that simplification we get: $\Large \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right) - \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \frac{{4xy - 9{x^2}y}}{{12{x^3}{y^2}}}$

87. Michele_Laino

now we can factor out, at numerator and at denominator the quantity xy, so we get: $\Large \begin{gathered} \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right) - \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \hfill \\ \hfill \\ = \frac{{4xy - 9{x^2}y}}{{12{x^3}{y^2}}} = \frac{{xy\left( {4 - 3x} \right)}}{{xy\left( {12{x^2}y} \right)}} \hfill \\ \end{gathered}$

88. Michele_Laino

am I right?

89. anonymous

yea

90. Michele_Laino

ok! Now we can cancel out xy, then we get: $\Large \begin{gathered} \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right) - \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \hfill \\ \hfill \\ = \frac{{4xy - 9{x^2}y}}{{12{x^3}{y^2}}} = \frac{{xy\left( {4 - 3x} \right)}}{{xy\left( {12{x^2}y} \right)}} = \hfill \\ \hfill \\ = \frac{{4 - 3x}}{{12{x^2}y}} \hfill \\ \end{gathered}$

91. anonymous

and is that it?

92. Michele_Laino

yes! we have finished!

93. anonymous

yey thank you!!!!!!

94. Michele_Laino

thank you!! :)

95. Michele_Laino

oops..there is a typo: $\Large \begin{gathered} \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right) - \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \hfill \\ \hfill \\ = \frac{{4xy - 9{x^2}y}}{{12{x^3}{y^2}}} = \frac{{xy\left( {4 - 9x} \right)}}{{xy\left( {12{x^2}y} \right)}} = \hfill \\ \hfill \\ = \frac{{4 - 9x}}{{12{x^2}y}} \hfill \\ \end{gathered}$

96. anonymous

ok got u thanks !!!

97. Michele_Laino

thanks! again! :)

98. anonymous

why u thanking me i sould be thanking u