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anonymous

  • 5 years ago

how do you solve this: 2x-3 = (3+x)/2

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  1. anonymous
    • 5 years ago
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    multiply by 2, subtract an x and add a 3

  2. anonymous
    • 5 years ago
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    thank you

  3. anonymous
    • 5 years ago
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    x=3

  4. anonymous
    • 5 years ago
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    i need the process, not the answer which i already know

  5. anonymous
    • 5 years ago
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    I don't think so

  6. anonymous
    • 5 years ago
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    The point is, you want to get all your x terms on one side and your non-variable terms on the other

  7. anonymous
    • 5 years ago
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    \[2*(2x-3)=[(3+x)/2]*2\]

  8. anonymous
    • 5 years ago
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    then it goes to : 4x-6=3+x

  9. anonymous
    • 5 years ago
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    thank you so much, i understand now

  10. anonymous
    • 5 years ago
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    the get x by itself...so subtract -x to the left so next step: 3x=9

  11. anonymous
    • 5 years ago
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    do i get a medal :)

  12. anonymous
    • 5 years ago
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    yes! but i don't know how to give one

  13. anonymous
    • 5 years ago
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    i need help with this one too please: 2x-1= 1/3(5-3x) + 4

  14. anonymous
    • 5 years ago
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    Ok, well why don't you tell us how to start this one

  15. anonymous
    • 5 years ago
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    Oh, and is it \[\frac{1}{3(5-3x)}\] or \[\frac{1}{3}(5-3x)\]

  16. anonymous
    • 5 years ago
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    the second one

  17. anonymous
    • 5 years ago
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    Ok. So which part do you want to tackle first? (There's lots of different paths to the same answer)

  18. anonymous
    • 5 years ago
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    Overall goal is to get all the x terms on one side and the non-x terms on the other side

  19. anonymous
    • 5 years ago
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    You tell me what to do, and I'll write the new equation.

  20. anonymous
    • 5 years ago
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    \[2x-1= \frac{1}{3}(5-3x) + 4\]

  21. anonymous
    • 5 years ago
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    i don't know how to start that's why i'm asking

  22. anonymous
    • 5 years ago
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    That's ok. Just pick a term that is in the wrong place

  23. anonymous
    • 5 years ago
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    We want everything with an x on the left, and everything without an x on the right

  24. anonymous
    • 5 years ago
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    Pick a term, any term.

  25. anonymous
    • 5 years ago
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    3x

  26. anonymous
    • 5 years ago
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    Ah, that term is a little tricky because it's part of a factor in a product. We will need to expand that product out first before we can work with that term directly

  27. anonymous
    • 5 years ago
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    But that's ok!

  28. anonymous
    • 5 years ago
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    That just means we need to distribute the 1/3 to each of the terms in the other factor

  29. anonymous
    • 5 years ago
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    So what do we get when we distribute that 1/3?

  30. anonymous
    • 5 years ago
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    3-9x?

  31. anonymous
    • 5 years ago
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    no. we have to multiply each term in the left factor by 1/3 \[\frac{1}{3}(5-3x) = \frac{1}{3}*5 - \frac{1}{3}*3x\]

  32. anonymous
    • 5 years ago
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    Does that make sense? So what do we have when we simplify that product?

  33. anonymous
    • 5 years ago
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    i'm not sure...

  34. anonymous
    • 5 years ago
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    a(b+c) = ab + ac Right? Basic multiplicative distribution

  35. anonymous
    • 5 years ago
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    ok so would it be 5/3 -x?

  36. anonymous
    • 5 years ago
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    yes

  37. anonymous
    • 5 years ago
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    So now we have: \[2x-1= \frac{5}{3}-x + 4\]

  38. anonymous
    • 5 years ago
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    so the whole thing would be : 2x-1 = 5/3-x+4?

  39. anonymous
    • 5 years ago
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    Yes

  40. anonymous
    • 5 years ago
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    Can you solve from there? or do you need more help with it?

  41. anonymous
    • 5 years ago
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    maybe a little more

  42. anonymous
    • 5 years ago
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    Ok, so pick a term that's in the wrong place

  43. anonymous
    • 5 years ago
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    how would i combine 5/3 + 4 that's the next step right?

  44. anonymous
    • 5 years ago
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    You can certainly do that yes.

  45. anonymous
    • 5 years ago
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    There is no 'next step'. There are a lot of ways to do it. If you want to do that next, go for it =)

  46. anonymous
    • 5 years ago
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    ok so how do you do it?

  47. anonymous
    • 5 years ago
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    How do you add \(\frac{5}{3} + 4\)

  48. anonymous
    • 5 years ago
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    You have to change the denominator of the 4 to match the 3 in the 5/3

  49. anonymous
    • 5 years ago
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    We do this using the trick of multiplying by 1.

  50. anonymous
    • 5 years ago
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    \(4 \times 1 = ?\)

  51. anonymous
    • 5 years ago
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    4

  52. anonymous
    • 5 years ago
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    Ok, so if we multiply by 1 we won't have changed the value, right?

  53. anonymous
    • 5 years ago
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    right

  54. anonymous
    • 5 years ago
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    Ok, so what is \(\frac{3}{3}\)

  55. anonymous
    • 5 years ago
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    1

  56. anonymous
    • 5 years ago
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    So you're saying then that if I multiply 4 by 3/3 I will still have 4?

  57. anonymous
    • 5 years ago
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    \[4\times \frac{3}{3} = \frac{4\times 3}{3} = \frac{12}{3}\]

  58. anonymous
    • 5 years ago
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    And 12/3 does equal 4, but it's in a different form that is now easy to add to 5/3

  59. anonymous
    • 5 years ago
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    Did you follow that?

  60. anonymous
    • 5 years ago
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    yes so the answer would be 17/3

  61. anonymous
    • 5 years ago
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    Well, the sum of those two terms would be 17/3 yes

  62. anonymous
    • 5 years ago
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    So we have: \[2x-1= \frac{17}{3}-x\]

  63. anonymous
    • 5 years ago
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    then irt would be 3x-1= 17/3

  64. anonymous
    • 5 years ago
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    Certainly

  65. anonymous
    • 5 years ago
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    it would be 3x = 1 17/3?

  66. anonymous
    • 5 years ago
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    yeah, but I dun like mixed numbers

  67. anonymous
    • 5 years ago
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    Can you do the same trick to the 1 we did to the 4 to make it easy to add to the 17/3

  68. anonymous
    • 5 years ago
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    isn't it 20/9?

  69. anonymous
    • 5 years ago
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    x = 20/9 is correct yes.

  70. anonymous
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    you don't divide that by 3?

  71. anonymous
    • 5 years ago
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    Huh? we did already

  72. anonymous
    • 5 years ago
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    It was 3x = 20/3

  73. anonymous
    • 5 years ago
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    oh ok i thought it was 3x = 1 17/3? and then you divide by 3 on both sides to get x

  74. anonymous
    • 5 years ago
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    \(3x = 1+\frac{17}{3}\)\[\implies 3x = 1\times \frac{3}{3} + \frac{17}{3}\]\[\implies 3x = \frac{1\times 3}{3} + \frac{17}{3}\]\[\implies 3x = \frac{3}{3} + \frac{17}{3}\]\[\implies 3x = \frac{20}{3}\]

  75. anonymous
    • 5 years ago
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    Then you divide by 3 on both sides

  76. anonymous
    • 5 years ago
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    Follow all that?

  77. anonymous
    • 5 years ago
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    yes thanks

  78. anonymous
    • 5 years ago
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    but how do you divide 3 on both sides, that what i want to know

  79. anonymous
    • 5 years ago
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    dividing by 3 is the same as multiplying by 1/3

  80. anonymous
    • 5 years ago
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    wouldn't the answer be different?

  81. anonymous
    • 5 years ago
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    nope. \[3x \times \frac{1}{3} = \frac{20}{3} \times \frac{1}{3}\] \[x = \frac{20}{9} \] \[3x \div 3 = \frac{20}{3} \div 3\] \[x = 6.\bar{6}\bar{6} \div 3 = 2.\bar{2}\bar{2} = \frac{20}{9}\]

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