## Firejay5 Group Title Simplify Each Expression. Show work and explain it. 32. m over m^2 - 4 + 2 over 3m + 6 one year ago one year ago

1. Firejay5

Well on #32, I am having trouble multiplying the denominators to make them the same

2. wio

"m over m^2 - 4 + 2 over 3m + 6" $\frac{m}{m^2-4}+\frac{2}{3m+6}$

3. Firejay5

why

4. wio

Just use this: $\frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd}$

5. wio

I think the + is separating two fractions.

6. Firejay5

yes

7. Firejay5

yea

8. Firejay5

can you do it with me and then I'll do 33.) my self

9. Firejay5

what do you mean?

10. Firejay5

Equation editor

11. Firejay5

can we get back to the problem

12. Firejay5

or not @Hero

13. Firejay5

the example he gave isn't going nowhere for me

14. wio

let $$b=m^2-4$$ and $$d=3m+6$$

15. wio

$\frac{m}{m^2-4}+\frac{2}{3m+6}=\frac{(m)(3m+6)+(2)(m^2-4)}{(m^2-4)(3m+6)}$

16. wio

@Firejay5 Do you see the pattern?

17. Firejay5

Yes? Kind of how did you get that

18. Firejay5

I need full out steps, like getting the denominators equal, etc.

19. Firejay5

@Mertsj could you help with that

20. mathslover

hmn . $\frac{m}{m^2-4}+\frac{2}{3m+6}$ Well I think that it would be Ok to factorize both denominators

21. Mertsj

Yes. mathslover will be able to show you.

22. Firejay5

@Mertsj you sure

23. mathslover

$$\color{blue}{m^2-4}$$ is in the form of $$a^2-b^2$$ $$a^2-b^2 = (a+b)(a-b)$$ Use the above stated identity to factorize $$m^2-4$$ .

24. Mertsj

yes

25. mathslover

Can you factor $$m^2-4$$ @Firejay5 ?

26. Firejay5

yes

27. mathslover

Can you tell me what would be the factorized form for $$x^2-4$$ ?

28. Firejay5

(x - 2) & (x + 2)

29. mathslover

Great! So $$x^2-4 = \color{blue}{(x+2)(x-2)}$$ Now. we have the complete question as : $$\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{3m+6}$$

30. mathslover

Now take 3 common from 3m + 6. What you get?

31. Firejay5

I did the other factorization as well

32. Firejay5

3(m + 2)

33. mathslover

Great! $$\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{\color{orange}{3(m+2)}}$$. Is what we have now.

34. mathslover

Now take LCM .

35. Firejay5

m + 2

36. mathslover

Check it again

37. Firejay5

and m - 2

38. mathslover

LCM of $$\cfrac{1}{a+b} + \cfrac{1}{2(a-b)}$$ (denominators) is (a+b)(a-b)2

39. mathslover

Similarly do that there. we have $$(x+2)(x-2) \quad\textbf{and}\quad 3(m+2)$$ as the denominators

40. Firejay5

it would be 3(m - 2) (m + 2)

41. mathslover

Yeah . and sorry there for using x on the place of m

42. mathslover

So I have : 3(m-2)(m+2) as LCM can you solve the fraction :

43. Firejay5

so what would it be entered in equation editor

44. mathslover

I didn't get you,, please explain!

45. Firejay5

type the problem in fraction form of what we have left

46. mathslover

hmn well it would be hard to say that I have : $$\cfrac{\textbf{yet to come}}{3(m+2)(m-2)}$$

47. mathslover

we have to solve numerator now

48. mathslover

can u do that?

49. Firejay5

wouldn't the numerator m + 2

50. mathslover

No! See : $$\cfrac{1}{(a+b)} + \cfrac{2}{(a-b)} = \cfrac{1(a-b) + 2(a+b)}{(a-b)(a+b)}$$

51. Firejay5

how does it work, you should + across

52. mathslover

|dw:1364264397027:dw|

53. Firejay5

doesn't make sense

54. mathslover

See i have LCM as 3(m+2)(m-2) I divided LCM by the first fraction's denominator that is (m+2)(m-2) to get 3 . Now I multiplied this 3 by the first fractions' numerator that is m

55. mathslover

getting it?

56. wio

Wow, finding the LCM is makes it complicated.

57. Firejay5

????

58. mathslover

Similarly for the second one.. I have LCM as 3(m+2)(m-2) I divided this LCM by the denominator of the second fraction 3(m+2) $$\cfrac{\cancel{3(m+2)} (m-2)}{\cancel{3(m+2)} }$$

59. mathslover

Does that mean you don't get it @Firejay5 ?

60. Firejay5

kind of

61. Firejay5

are the denominators the same

62. mathslover

No.

63. mathslover

|dw:1364264678904:dw|

64. Firejay5

8/15

65. mathslover

But how?

66. Firejay5

whatever you multiply to the bottom, you must do to the top

67. Firejay5

work the problem a whole different way

68. mathslover

|dw:1364264834069:dw|

69. Firejay5

2(a - c) or 2(a + c)

70. mathslover

The LCM would be : 2(a+c)(a-c)

71. Firejay5

let's work the problem a different way and not factorize, and get the denominators the same

72. mathslover

Yes , that would be easy for u.

73. mathslover

Well which denominator you want to be changed

74. Firejay5

You're way is not what we were taught in class and is confusing me. 3m + 6

75. mathslover

Sorry, ok let us change 3(m+2)

76. mathslover

Would you try that for once?

77. Firejay5

what

78. mathslover

to change the denominator ?

79. Firejay5

I have been trying

80. mathslover

ok! I have first denominator as (m+2)(m-2) and other one as 3(m+2) I want to make 3(m+2) = (m+2)(m-2) For that I think I should multiply 3(m+2) by $$\cfrac{(m-2)}{3}$$

81. Firejay5

where did you get m - 2 over 3

82. Firejay5

don't jump ahead, it feels like you are just picking random numbers and variable from the problem

83. Mertsj

Are you all clear now, Firejay?

84. Firejay5

kind of would you explain it the same way

85. Mertsj

Do you want to talk about the original problem or what?

86. Firejay5

It feels like I am being taught a whole different then I learned it in class

87. Firejay5

different way

88. Mertsj

I just want to make sure we are talking about the right problem. Is this it:

89. Mertsj

|dw:1364265923023:dw|

90. Firejay5

yes! :D

91. Mertsj

|dw:1364265995425:dw|

92. Mertsj

When you add fractions you need to have the denominators the same. So we have to build the common denominator.

93. Firejay5

@mathslover you did good though! :D

94. Mertsj

For the first denominator we need the factors m-2 and m+2. For the second denominator we need the factors 3 and m+2. But we already have m+2. So the 3 factors we need are 3(m-2)(m+2)

95. Mertsj

|dw:1364266190884:dw|

96. Mertsj

Are you with me so far?

97. Mertsj

|dw:1364266301603:dw|

98. Mertsj

And finally: |dw:1364266416514:dw|

99. Firejay5

does it matter which order the m - 2 or m + 2 goes

100. Firejay5

uh???

101. Firejay5

I put 5m - 4 over 3(m + 2)(m - 2)

102. Mertsj

It doesn't matter.

103. Firejay5

How would you type 33.) y over y + 3 (-) 6y over y^2 - 9 in Wolframalpha

104. Mertsj

$\frac{y}{y+3}-\frac{6y}{y^2-9}$

105. Firejay5

yea

106. Mertsj

For wolf: y/(y+3)-(6y)/(y^2-9)

107. Firejay5

which of the alternative forms are the answer

108. Mertsj

|dw:1364267512777:dw|

109. Mertsj

That's the abbreviated version.

110. Firejay5

is it the first, second, or third one

111. Mertsj

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