## anonymous 5 years ago Could someone tell me if my answer was correct? I will attach the equation and available answers in a comment! Its just a algebra question but im evidently unable to focus rite now >.<

1. anonymous

here :)

2. anonymous

Which one is the question?

3. anonymous

the one that says "ok" ^_^

4. anonymous

I see.

5. anonymous

The process is confusing me im pretty sure this is one of the lessons that i missed :P

6. anonymous

So the question is to simplify: ${4 \over m+2}+{2 \over m-2}?$

7. anonymous

The m + 2 and m - 2 is confusing me :P And it says to Add and Simplify.

8. anonymous

Good. First, what is the LCM of (m+2) and (m-2).

9. anonymous

Thats the part that was confusing me ^_^ lol

10. anonymous

Haha I see. Do (m+2) and (m-2) have any common factors?

11. anonymous

Yep lol >.< i think urghh i should of paid attention :P and i just didnt know how to write it bc of the - and + signs >.< lol

12. anonymous

Well they don't have a common factor, and because of that the LCM of them is their multiplication.

13. anonymous

Ahhhhhh alright i get it so far. ^_^

14. anonymous

so (m+2)(m-2)=m^2-4. That's would be our LCM.

15. anonymous

That's good. Now, we have to apply this to our equation.

16. anonymous

expression*

17. anonymous

ohhhh haha wow ya bc since they dont have a common factors u'd multiply them together. alright.

18. anonymous

Exactly.

19. anonymous

That means, we will multiply the first term (top and bottom) by (m-2). and multiply the second term (top and bottom by (m+2).

20. anonymous

Does that make sense?

21. anonymous

Yes so far so good so then the answer would be the "ya" one? lol like my labels? :D

22. anonymous

>.< i think... :P

23. anonymous

${4 \over m+2}+{2 \over m-2}={4(m-2) \over m^2-4}+{2(m+2) \over m^2-4}={4m-8+2m+4 \over m^2-4}={6m-4 \over m^2-4}$

24. anonymous

OR as in the choices: ${6m-4 \over (m+2)(m-2)}$

25. anonymous

WOOOOOOOOOOOOO!!!! so i was rite huh?!?! YAY YAY YAY!!

26. anonymous

lol ^_^

27. anonymous

Haha Yeah you were :)

28. anonymous

:) Thanks, helped my bunches! :D

29. anonymous

np

30. anonymous

Oh! and do you think you could help me on part of this other question? :) plesss? lol

31. anonymous

This is it. do u think u will help meh? :)

32. anonymous

so ok is the original expression?

33. anonymous

You make strange names? :P

34. anonymous

My question is about the denominators again so would there be common factors? >.< i got confused again :P lol and ya i am making strange names ^^

35. anonymous

The same first question again, do they have common factors?

36. anonymous

And yea this is the original expression .. pretty sure lol

37. anonymous

and i thought they did >.< like since 3x can go into 6x.. but then i got lost arghh

38. anonymous

Okay can you factor any of the two denominators?

39. anonymous

No i dont think so

40. anonymous

Wait which version of factoring? >.<

41. anonymous

3x+5 can't be factored.. (6x+10)=2(3x+5).. right?

42. anonymous

Right.

43. anonymous

Okay now your LCD should be each one of the factor to be taken only one. the factors we have are (3x+5) and 2. so our LCD would 2(3x+5).

44. anonymous

I don't think that was clear enough, was it?

45. anonymous

Not really im sitting here repeating it over in my head going wth.... ^_^

46. anonymous

LOL

47. anonymous

Well, you can see that (3x+5) occur in both denominators. right?

48. anonymous

after factorization*

49. anonymous

Yep i see that.

50. anonymous

${2 \over 3x+5}-{1 \over 6x+10}={2 \over 3x+5}-{1 \over 2(3x+10)}$ Right?

51. anonymous

Now list all factors in both denominators.

52. anonymous

I mean 2(3x+5) :)

53. anonymous

Ahhhh so then i see why you would do that. But couldnt you just multiply the 3x + 5 expression by 2 and then subtract?

54. anonymous

the two equations with the same denominator i mean ^^

55. anonymous

Yes you could, but then you have to simplify after that.

56. anonymous

that wouldn't be the simplest form.

57. anonymous

Ahhh rite, ^_^ ok then go on lol

58. anonymous

Well you're saying that you multiply the (3x+5) by 2, that's right. And that's what I was trying to say when I said that 2(3x+5) is the LCD. So then we keep the 6x+10 and multiply the (3x+5) term (top and bottom) by 2.

59. anonymous

:)

60. anonymous

ohh so the answer would be B haha wow i rly need to stop doubting myself argh :P

61. anonymous

LOL thanks for the help! AGAIN ^_^

62. anonymous

But you know why we can do that. That's because the denominator of the second term is nothing but 2 times the denominator of the first term.

63. Yuki

AnwarA, how did you make the rational expression look like that ?

64. anonymous

It's 2(3x+5).. just typo :P

65. Yuki

all I can do is${a+bi}/{10a-c}$

66. Yuki

how did you make it go under a long bar ?

67. anonymous

lol think person means like how do u ya long bar ^^

68. anonymous

Oh.. {a+bi over 10a=c}

69. anonymous

And hello yuki.

70. anonymous

:)

71. anonymous

>.< i will never understand computer tech/ math peoples smartness :P

72. anonymous

haha. I think I am the only one who does it this way :P

73. Yuki

$\sqrt(b^2-4ac) \over 2a$

74. Yuki

that is sooo cooooool !! I am goint to use it from now on, thanks!

75. anonymous

LOL obviously not anymore!! :D and i see someone ik viewing the question but he isnt in our meeting place!! >:P lol (inside joke cant rly explain it ^^)

76. anonymous

${\sqrt{b^2-4ac} \over 2a}$ :P

77. Yuki

can you do "plus or minus" as well ?

78. anonymous

Haha

79. anonymous

>.< smart people arg lol

80. anonymous

Yep anything

81. anonymous

${-b \pm \sqrt{b^2-4ac} \over 2a}$

82. Yuki

how did you make the sqrt top part long ?! I have so much to learn lol

83. anonymous

Haha try that yourself :P

84. Yuki

is there a place that have all of that written down ?

85. anonymous

*sigh* so much to learn to little time lmao im jk :)

86. anonymous

just use {}, with anything that you want to make long :)

87. Yuki

$\sqrt{10x+b}$

88. anonymous

Hah, i think its funnay how i have no idea how the heck u guys are doing that ^^

89. anonymous

$e^{ANWAR}={1 \over \sqrt{yuki}^{dolly}}$ :P that does not mean anything by the way

90. Yuki

$\int\limits _{a,b}_{b,a}$

91. anonymous

lol im above yuki hee hee!!!!

92. anonymous

I am in the top, that's what matters :D

93. Yuki

$\ln(\ln (x)) = 0$

94. anonymous

:P looks more like im the glue that holds everything together mwuahahaha!!!

95. anonymous

take ln of both sides to bring me down :(

96. Yuki

$+,- \pm$

97. Yuki

Ha ! found it lo

98. anonymous

You're learning :P

99. Yuki

$\cos(a \pm b) = \cos(a)\cos(b) \mp \sin(a)\sin(b)$

100. anonymous

im not ^_^ LOL

101. anonymous

You want to?

102. anonymous

But you can't use your weird names there :P

103. anonymous

yuki, whenever you use these things, rmr to thank me first

104. anonymous

awww yep sure the person i am waiting for isnt talking so sure :) lol

105. Yuki

$\int\limits_{a,b \int\limits_{1}^{2}}$

106. anonymous

Haha

107. anonymous

whom are you waiting for? xD

108. Yuki

I definitely will AnwarA

109. anonymous

his name starts with an I and he is rlyyyy sweet and hot and all tht jaz oh and did i mention it was u ? xD lmao

110. anonymous

haha anwar is all like wat the hell.... LOL

111. anonymous

Thanks buddy!! I was just kidding. I did nothing, anyone who knows that would help do that. :)

112. anonymous

@dolly: LOL

113. Yuki

${a+bi \over a-bi} *{a+bi \over a-bi}$

114. anonymous

^_^

115. anonymous

why do you think I was "all like what the hell"? :)

116. anonymous

haha to what i said lol jeezus anwar pay attention!! >:D im jk :)

117. anonymous

Haha, I read that and I wasn't "all like what the hell" at all :P

118. anonymous

yuki, what grade are you at?

119. anonymous

LOL then i was wrong i thought u would be ^_^ lol and prolly college or high school senior!! lol

120. anonymous

I think wherever I need info about anyone in this site, I should just come straight to you. You got it all :O

121. anonymous

Do you work for the CIA?

122. anonymous

haha its easy i just notice the kind of math they know how to do, the way they type.. etc :) nm too it :D

123. anonymous

so you are not here to gather informations about us for some agencies?

124. anonymous

Uhh, no. lol That would only be extremely creepy lol

125. anonymous

:) Anyway, I am off to bed. Yuki would be more of help than I am, specially that he now knows how to do the equations thing :D.

126. anonymous

Bye all!!

127. anonymous

LOL okie dokie nite buddaaayyyy!!! :)

128. Yuki

sigma