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aspenfields

  • one year ago

how do i solve the system of equations? 2x + 4y + 3z = 6 3x + 5y + 6z =3 2x + 3y + 4z = 8

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  1. beketso
    • one year ago
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    are you familiar with matrices

  2. aspenfields
    • one year ago
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    kinda

  3. beketso
    • one year ago
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    well you can use the gaussian elimination

  4. aspenfields
    • one year ago
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    ok

  5. beketso
    • one year ago
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    meaning you use the elementary row operations

  6. aspenfields
    • one year ago
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    yea

  7. beketso
    • one year ago
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    2 4 3 | 6 3 5 6 | 3 2 3 4 | 8 subtract row 1 from row 3 and tell me what u get

  8. aspenfields
    • one year ago
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    ok one sec

  9. aspenfields
    • one year ago
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    0 -1 1 |2 is that right?

  10. beketso
    • one year ago
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    yeah.now divide row 1 by 2

  11. aspenfields
    • one year ago
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    ok the answer i have above isnt row one is it or is row one in the problem still?

  12. beketso
    • one year ago
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    no.it is still row 3.row one is ur original problem 2 4 3 | 6

  13. aspenfields
    • one year ago
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    ok

  14. aspenfields
    • one year ago
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    2/3 4/5 1/2 |2 right?

  15. beketso
    • one year ago
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    it is actually 1 2 3/2 | 3 get it?

  16. aspenfields
    • one year ago
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    how?

  17. beketso
    • one year ago
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    well u take the first row.and u divide by 2.then it changes from 2 4 3 |6 to 1 2 3/2 | 3

  18. aspenfields
    • one year ago
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    ohh! u divid by two not thr 2nd row :/lol now i got that ok

  19. beketso
    • one year ago
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    then your rows are now like this 1 2 3/2 | 3 3 5 6 | 3 0 -1 1 | 2

  20. aspenfields
    • one year ago
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    yea

  21. beketso
    • one year ago
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    sorry. multipy row 1 by 3 the subtract it from row 2 the u have a reduced row 2

  22. aspenfields
    • one year ago
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    ok? one sec

  23. aspenfields
    • one year ago
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    when i subtract which number should be on top?

  24. beketso
    • one year ago
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    1 2 3/2 | 3 _ _ _ | _ 0 -1 1 | 2 the only r0w that changes is row 2

  25. aspenfields
    • one year ago
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    yea,ok i get that so row to is now 3 8 9/2 |-3

  26. beketso
    • one year ago
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    3-(3X3) = 0 That is how u do it.

  27. aspenfields
    • one year ago
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    what?

  28. beketso
    • one year ago
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    meaning u take 1 from row1 and multiply by 3 the u subtract the answer from 3 in row 2 and u get ur as 0. do the same with the second term

  29. beketso
    • one year ago
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    then u get 1 2 3/2 | 3 0 -1 3/2 | -6 0 -1 1 | 2

  30. aspenfields
    • one year ago
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    ohh

  31. aspenfields
    • one year ago
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    ok then from there what do i do next ?

  32. beketso
    • one year ago
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    multiply row 2 by -1

  33. aspenfields
    • one year ago
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    ok

  34. aspenfields
    • one year ago
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    0 1 -3/2 |6

  35. beketso
    • one year ago
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    exactly

  36. aspenfields
    • one year ago
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    ok is this the answer or is the more?

  37. beketso
    • one year ago
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    1 2 3/2 | 3 0 1 -3/2 | 6 0 -1 1 |2 in row 1 u must eliminate 2 and in row 3 must eliminate -1.

  38. aspenfields
    • one year ago
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    why?

  39. beketso
    • one year ago
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    add row 3 and row 2 to get 1 2 3/2 |3 0 1 -3/2 |6 0 0 1/2 |-4 and u jst eliminated 1 from row 3

  40. aspenfields
    • one year ago
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    ok

  41. beketso
    • one year ago
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    sorry u jst eliminated -1 from row 3

  42. beketso
    • one year ago
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    then now we eliminate 2 from row 1

  43. aspenfields
    • one year ago
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    i figured u ment that

  44. aspenfields
    • one year ago
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    :)

  45. beketso
    • one year ago
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    do this row1-2(row2)

  46. beketso
    • one year ago
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    then u get 1 0 9/2 |-9 0 1 -3/2 | 6 0 0 1/2 |-4

  47. beketso
    • one year ago
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    getting this?

  48. aspenfields
    • one year ago
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    yea

  49. beketso
    • one year ago
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    basically u want to get to a point were u have 1 0 0 | ? 0 1 0 | ? 0 0 1 | ?

  50. beketso
    • one year ago
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    now divide row 3 by 1/2 (half)

  51. aspenfields
    • one year ago
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    ohh ok

  52. aspenfields
    • one year ago
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    1 |-2

  53. beketso
    • one year ago
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    what is 4 divided by half

  54. aspenfields
    • one year ago
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    oh 8

  55. beketso
    • one year ago
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    then u have 1 0 9/2 | -9 0 1 -3/2 | 6 0 0 1 | -8 what do u think u should do to eliminate 9/2?

  56. aspenfields
    • one year ago
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    2/9?

  57. beketso
    • one year ago
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    what about 2/9?

  58. aspenfields
    • one year ago
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    u mulitply?

  59. beketso
    • one year ago
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    remember you want it to be 0

  60. aspenfields
    • one year ago
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    on then subtract 9/2?

  61. beketso
    • one year ago
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    exactly

  62. aspenfields
    • one year ago
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    ok :)

  63. aspenfields
    • one year ago
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    what bout the numbers on the end?

  64. beketso
    • one year ago
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    so mentally multiply row 3 by 9/2

  65. beketso
    • one year ago
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    and then subtract it from row 1

  66. beketso
    • one year ago
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    what do u get for row1?

  67. aspenfields
    • one year ago
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    1 0 0 |1?

  68. beketso
    • one year ago
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    actually 1 0 0 |27

  69. aspenfields
    • one year ago
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    oh

  70. beketso
    • one year ago
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    now u got 1 0 0 | 27 0 1 -3/2 |6 0 0 1 |-8

  71. beketso
    • one year ago
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    u must now eliminate -3/2.how do u think u do that?

  72. aspenfields
    • one year ago
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    yes

  73. beketso
    • one year ago
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    how do you think u can do it?

  74. aspenfields
    • one year ago
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    plus 3/2

  75. beketso
    • one year ago
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    sure.which row can u use to eliminate it?

  76. aspenfields
    • one year ago
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    3

  77. beketso
    • one year ago
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    good.how do u do that?

  78. aspenfields
    • one year ago
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    idk

  79. beketso
    • one year ago
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    multiply row3 by 3/2 and add it to row2

  80. aspenfields
    • one year ago
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    ok

  81. beketso
    • one year ago
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    that way 1 becomes 3/2 and you can add that to row 2.the -3/2 + 3/2 = 0 which eliminates 3/2

  82. aspenfields
    • one year ago
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    yea

  83. beketso
    • one year ago
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    then u have 1 0 0 |27 0 1 0 |-6 0 0 1 |-8 what is ur x,y and z?

  84. aspenfields
    • one year ago
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    27,-6,-8?

  85. beketso
    • one year ago
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    yes

  86. aspenfields
    • one year ago
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    ok

  87. beketso
    • one year ago
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    but i think i made a mistake with the numbers somewhere

  88. beketso
    • one year ago
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    should i paste the whole solution or can u figure out the mistake?

  89. aspenfields
    • one year ago
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    i can figure it out now

  90. beketso
    • one year ago
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    ok.sure

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