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anonymous

  • one year ago

Algebra Question

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  1. anonymous
    • one year ago
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    If \[\frac{ a }{ 2 } + \frac{ b }{ 2 } = 3 \] what is the value of 2a +2b?

  2. anonymous
    • one year ago
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    I haven't really been able to get anywhere on it, any form of help will help :D

  3. anonymous
    • one year ago
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    u still there

  4. anonymous
    • one year ago
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    what do u think

  5. anonymous
    • one year ago
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    yeah im back sorry

  6. anonymous
    • one year ago
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    @slade well I don't really have any idea on how to start

  7. freckles
    • one year ago
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    I dare you to multiply both sides by 4

  8. anonymous
    • one year ago
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    thats how u do it right there

  9. freckles
    • one year ago
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    double dog dare you

  10. anonymous
    • one year ago
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    what would that do? um okaay 3 x 4 = 12 on the right but i don't know how to multiply the fractions :/

  11. freckles
    • one year ago
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    do you know what 4/2=?

  12. anonymous
    • one year ago
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    yes 2

  13. freckles
    • one year ago
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    \[4(\frac{a}{2}+\frac{b}{2}) \\ =\frac{4}{2}a+\frac{4}{2}b=? \]

  14. anonymous
    • one year ago
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    a/2 and b/2 x 4?

  15. anonymous
    • one year ago
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    yes

  16. anonymous
    • one year ago
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    ah gotcha

  17. anonymous
    • one year ago
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    2a + 2b is what i get! and that means it eqauls 3 !!

  18. freckles
    • one year ago
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    no

  19. freckles
    • one year ago
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    3(4)

  20. anonymous
    • one year ago
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    but what would make me multiply each side by 4?

  21. anonymous
    • one year ago
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    oh yeah 12

  22. freckles
    • one year ago
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    to solve for 2a+2b

  23. anonymous
    • one year ago
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    I like your approach @freckles

  24. anonymous
    • one year ago
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    2 * 4 =?

  25. anonymous
    • one year ago
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    I dare you all to learn math :-D double dog dare at that :-D

  26. anonymous
    • one year ago
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    to solve yes but where did you think of 4 why not 3? like is there a specific reason or thats just how algebra works? common factor what was it?

  27. freckles
    • one year ago
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    well 4(a/2+b/2) would give me what I want which is 2a+2b

  28. anonymous
    • one year ago
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    i'm trying nixy lol

  29. freckles
    • one year ago
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    but if I multiply one side by 4 I have to multiply the other side by 4

  30. anonymous
    • one year ago
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    ah yes that is true that is how to get the end result

  31. freckles
    • one year ago
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    do you want to give you a problem that is similar ?

  32. anonymous
    • one year ago
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    sure

  33. anonymous
    • one year ago
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    @kevinhunterwood69 it is hard work but if you keep at it, it will get easier :-D

  34. anonymous
    • one year ago
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    yeah i am

  35. freckles
    • one year ago
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    solve for 6a+9b given \[2a+3b=-1\]

  36. anonymous
    • one year ago
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    im out

  37. anonymous
    • one year ago
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    no one is listening to me

  38. anonymous
    • one year ago
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    lol there

  39. freckles
    • one year ago
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    goodbye @slade :(

  40. freckles
    • one year ago
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    @kevinhunterwood69 do you notice how to find the value of 6a+9b yet?

  41. anonymous
    • one year ago
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    okay so I did: solve for 6a + 9b given: 2a + 3b = -1 work: 2a + 3b = -1 3 (2a + 3b) = -1(3) 6a + 9b = -3 answer: -3

  42. anonymous
    • one year ago
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    was that correct?

  43. freckles
    • one year ago
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    omg brilliant

  44. anonymous
    • one year ago
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    yay :D

  45. anonymous
    • one year ago
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    math is fun when i can finally understand algebra

  46. freckles
    • one year ago
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    good job I bet if that was the problem you had you would have had no problems do you think the fractions throw you off ?

  47. anonymous
    • one year ago
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    yes def fractions have always been a struggle to understand

  48. anonymous
    • one year ago
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    square roots too but mostly fractions

  49. freckles
    • one year ago
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    try this one: \[\text{ If } a+\frac{b}{3}=4 \text{ then what is the value of } 21a+7b\] (this will has a fraction in it; let's see if it throws you off any this time )

  50. anonymous
    • one year ago
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    hmm

  51. anonymous
    • one year ago
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    if i move 3 to the other side will it be on top?

  52. anonymous
    • one year ago
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    actually wait i can't do that right ugh um if i move b let me see

  53. freckles
    • one year ago
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    so you want to multiply both sides 3 to remove the fraction you can do that 3a+b=12 but we still are not done

  54. anonymous
    • one year ago
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    all i know is im trying to get constants on one side and variables on the other right?

  55. anonymous
    • one year ago
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    sorry what

  56. anonymous
    • one year ago
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    how did you get 3a and b lost its 3? oh you moved the 3 from b next to a?

  57. freckles
    • one year ago
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    \[a+\frac{b}{3}=4 \\ \text{ if you don't like the fraction there you can always multiply both sides by } \\ \text{ what is \in the denominator of that fraction } \\ 3(a+\frac{b}{3})=3(4) \\ 3(a)+3(\frac{b}{3})=3(4) \\ 3a+b=12\]

  58. anonymous
    • one year ago
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    yes that explains it! thanks so much

  59. anonymous
    • one year ago
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    well 3a i understand actually haha 3(b/3) = b?

  60. freckles
    • one year ago
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    \[3(\frac{b}{3})=3 \cdot \frac{b}{3}=\frac{3}{1} \cdot \frac{b}{3}= \frac{3}{3} \cdot \frac{b}{1}=1 \cdot b =b \]

  61. freckles
    • one year ago
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    basically in summary 3/3=1 so you are left with 1*b which is just b

  62. freckles
    • one year ago
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    like in your problem we had above we had: \[\frac{a}{2}+\frac{b}{2}=3 \\ \text{ we could have cleared the fractions before finding what we wanted } \\ \text{ by multiplying both sides by 2} \\ 2(\frac{a}{2}+\frac{b}{2})=2(3) \\ 2 \frac{a}{2}+2 \frac{b}{2}=2(3) \\ a+b=6\]

  63. anonymous
    • one year ago
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    yes i follow up to the point 3/3 = b/1 why does b not become 3b? like 3/3 = 3b/1 ?

  64. freckles
    • one year ago
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    anything over itself is 1 (well except 0)

  65. freckles
    • one year ago
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    well 3/3 *b =b

  66. freckles
    • one year ago
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    first do you know that 3/3=1 ?

  67. anonymous
    • one year ago
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    yes i do

  68. anonymous
    • one year ago
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    no time for jokes lol

  69. freckles
    • one year ago
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    \[\frac{3}{3}=1 \\ \text{ so if I multiply both sides by } b \text{ we have } \\ \frac{3}{3}b=1 b\]

  70. anonymous
    • one year ago
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    yes

  71. freckles
    • one year ago
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    is this not answering your question?

  72. anonymous
    • one year ago
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    no it did

  73. anonymous
    • one year ago
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    so we were at 3a + b = 12

  74. anonymous
    • one year ago
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    and m trying to figure out what is 21a + 7b so

  75. freckles
    • one year ago
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    right and we want 21a+7b 's value

  76. anonymous
    • one year ago
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    yes

  77. freckles
    • one year ago
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    any if you multiply both sides by s_v__ you will get?

  78. freckles
    • one year ago
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    I didn't put the full word there

  79. anonymous
    • one year ago
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    ah seven

  80. anonymous
    • one year ago
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    wait a second

  81. anonymous
    • one year ago
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    then we multiply 12 by 7 that means :O

  82. freckles
    • one year ago
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    yep! :)

  83. freckles
    • one year ago
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    just so you know we could have figured out 21a+7b in one step

  84. freckles
    • one year ago
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    instead of two

  85. anonymous
    • one year ago
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    ya so 21a + 7b = 84

  86. freckles
    • one year ago
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    \[a+\frac{b}{3}=4 \\ \text{ we wanted } 21a+7b \\ \text{ so we could have just multiplied 21 on both sides \to \begin with } \\ 21a+21(\frac{b}{3})=21(4) \\ 21a+7b=84\]

  87. anonymous
    • one year ago
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    wow

  88. freckles
    • one year ago
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    anyways keep practicing and numbers will become easier to play with

  89. anonymous
    • one year ago
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    thanks for all the extra practice!

  90. freckles
    • one year ago
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    np :) these can get a lot funnier I promise \[\text{ assume } a \text{ and } b \text{ are positive } \\\text{ if } a^2+2ab+b^2=4 \text{ then what is } (a+b)^4 \text{ or what is } a+b \] I think they can even get funnier than that

  91. anonymous
    • one year ago
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    uh lol

  92. freckles
    • one year ago
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    \[a^2+2ab+b^2=(a+b)^2 \text{ is the trick there }\]

  93. anonymous
    • one year ago
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    ah

  94. anonymous
    • one year ago
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    welp i'll go back to this sat book

  95. freckles
    • one year ago
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    k k have fun :)

  96. anonymous
    • one year ago
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    and ponder on about the funny stuff in math lol

  97. anonymous
    • one year ago
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    thx frecks

  98. freckles
    • one year ago
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    np :)

  99. mathstudent55
    • one year ago
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    @freckles Great explanations!

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