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anonymous

  • one year ago

how do i find a solution to a system of equations represented by a matrix?

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  1. anonymous
    • one year ago
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    this specifically

  2. anonymous
    • one year ago
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    Well first you need to know how to read a matrix. Do you know how to read a matrix?

  3. anonymous
    • one year ago
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    no....

  4. anonymous
    • one year ago
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    im kind of clueless

  5. anonymous
    • one year ago
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    and im bad at asking for help

  6. anonymous
    • one year ago
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    |dw:1438709526900:dw|

  7. anonymous
    • one year ago
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    Okay I'm just writing the problem here so I can see it more easier. And it's okay. I'll write how to solve the problem step by step. Just tell me when you are lost.

  8. anonymous
    • one year ago
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    aw, thank you

  9. anonymous
    • one year ago
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    You're welcome! :)

  10. anonymous
    • one year ago
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    itll help alot if i understand this

  11. anonymous
    • one year ago
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    Alright I'll try the best I can in getting you to understand this. |dw:1438709667710:dw|

  12. anonymous
    • one year ago
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    First you need to know how to read the matrix. As you can see there is a column name called x y and z. Those numbers below the variables are going to be part of the equation.

  13. anonymous
    • one year ago
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    For example in the first row you see the numbers 1 2 and -2. When you put this into an equation you will use the variables above the first row to combine it as shown below:\[x+2y-2z\]The 11 will be after the equals sign since that line separating -2 and 11 technically means equals. So now the equation will look like this:\[x+2y-2z=11\]

  14. anonymous
    • one year ago
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    Did you get that so far?

  15. anonymous
    • one year ago
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    uuhm..

  16. anonymous
    • one year ago
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    oh, yeah okay i get it.

  17. anonymous
    • one year ago
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    since one would be just x, sorry takes me awhile to work things out.

  18. anonymous
    • one year ago
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    It's okay! Take your time! :)

  19. anonymous
    • one year ago
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    Do you know what the other two equations will be if you used my method that I showed you?

  20. anonymous
    • one year ago
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    5x+4y+7z=81?

  21. anonymous
    • one year ago
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    Yes!

  22. anonymous
    • one year ago
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    -4x+y-6z=-37

  23. anonymous
    • one year ago
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    Great job! Now those three equations are needed when you are trying to solve equations within a matrix.

  24. anonymous
    • one year ago
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    \[x+2y-2z=11\]\[5x+4y+7z=81\]\[-4x+y-6z=-37\]

  25. anonymous
    • one year ago
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    Now do you know how to solve for x's and y's?

  26. anonymous
    • one year ago
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    replace them with the numbers or..

  27. anonymous
    • one year ago
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    Not in this case since they didn't give us any numbers to put in for x y or z. So the first step is to pick any two sets of equations out of the three.

  28. anonymous
    • one year ago
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    Let's start with the first two equations.

  29. anonymous
    • one year ago
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    sorry, ah :(

  30. anonymous
    • one year ago
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    |dw:1438710552563:dw|

  31. anonymous
    • one year ago
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    And it's okay. That was a rhetorical question.

  32. anonymous
    • one year ago
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    Oh wait let me rewrite what I wrote on the side.

  33. anonymous
    • one year ago
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    Pick which variable to make common. For example the y's can be made common by multiplying the 2y by 2 and leaving the 4y. When you multiply 2y by 2 you can get 4y. 4y and 4y equal each other so now you can subtract the whole equation to get rid of y.

  34. anonymous
    • one year ago
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    Tell me if you get that part.

  35. anonymous
    • one year ago
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    hm..

  36. anonymous
    • one year ago
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    i do, but at the same time i guess i need to know why we are doing that.

  37. anonymous
    • one year ago
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    what is the reason for finding commons, is that like a big part of the equation?

  38. anonymous
    • one year ago
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    what is the end result supposed to be? like, is the Y now 1? bleh...

  39. anonymous
    • one year ago
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    We are doing that so that we can get rid of one term and just concentrate on finding x and z. We will find y at the end.

  40. anonymous
    • one year ago
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    why are we doing that?

  41. anonymous
    • one year ago
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    In order to find the solution to this matrix we need to find what x y and z equals. So we need to break it down by finding one variable at a time.

  42. anonymous
    • one year ago
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    Oh alright. so whenever im doing this i have to find a way to get rid of a variable?

  43. anonymous
    • one year ago
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    Yes.

  44. anonymous
    • one year ago
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    okay, so, like in this equation for example: |dw:1438711521979:dw|

  45. anonymous
    • one year ago
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    just as an example

  46. anonymous
    • one year ago
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    they have to be in the same row to be able to get rid of them right? the 9 and -9 cant cancel each other out

  47. anonymous
    • one year ago
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    I'm sorry but where did 31 -25 and 28 come from? It's the last column. And no you can't cancel out the 9 and -9 because one is a y and the other is a z. Remember the 2nd column are y's and the 3rd column are z's.

  48. anonymous
    • one year ago
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    |dw:1438711724574:dw|

  49. anonymous
    • one year ago
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    its an example problem of one i did and failed.

  50. anonymous
    • one year ago
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    totally different problem sorry just trying to see where this applies to other problems too

  51. anonymous
    • one year ago
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    You understood more than before at least :) And oh ok. I'll help you out with those if you don't understand still.

  52. anonymous
    • one year ago
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    thank you, youre a saint ;-;

  53. anonymous
    • one year ago
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    okay, so lets finish this problem first

  54. anonymous
    • one year ago
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    Lol you're welcome! And okay. We have one equation: -3x-11z=-59 after canceling the y. Now when you look at the 3 equations from earlier, you can see that we need an equation with just an x and z and no y but we don't have that. So we need to repeat the step in finding another 2 equations to cancel out y.

  55. anonymous
    • one year ago
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    did you subtract?

  56. anonymous
    • one year ago
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    im lost again

  57. anonymous
    • one year ago
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    -3x-11z=-59 came from the drawing I drew earlier. Take a look at that and tell me if you get that.

  58. anonymous
    • one year ago
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    where did the other two come from?

  59. anonymous
    • one year ago
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    It came from the three equations we wrote from the matrix.

  60. anonymous
    • one year ago
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    |dw:1438712602958:dw| gosh im so sorry, im not sure where the two came from. i wrote down the equation again, to see if i can do what youre doing

  61. anonymous
    • one year ago
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    Oh lol okay.

  62. anonymous
    • one year ago
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    https://join.me/592-430-542 im sharing my screen with you, is that alright??

  63. anonymous
    • one year ago
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    I'm pretty freaked out by that screen. Lol sorry but I don't feel like it's safe. No offense to you or anything. I just feel like it's a hack since that kind of screen popped up when my uncle's computer got hacked.

  64. anonymous
    • one year ago
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    pfft, im not trying to hack anyone and learn math at the same time but thats understandable. i usually use joinme with my friends when im drawin and stuff

  65. anonymous
    • one year ago
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    Oh okay. I'm really sorry about that! Are you able to draw here?

  66. anonymous
    • one year ago
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    We're almost done anyways.

  67. anonymous
    • one year ago
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    |dw:1438713360885:dw|

  68. anonymous
    • one year ago
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    alright, sorry i scared you haha

  69. anonymous
    • one year ago
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    the brushes on here kind of suck there for i kind of suck

  70. anonymous
    • one year ago
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    It's okay you didn't know! :) And at least you draw better than me XD

  71. anonymous
    • one year ago
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    \[x+2y-2z=11\]\[5x+4y+7z=81\]\[-4x+y-6z=-37\]*Just rewriting this so I don't have to keep scrolling up.*

  72. anonymous
    • one year ago
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    i changed the background so its less menacing lol

  73. anonymous
    • one year ago
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    but anyways, MATH.

  74. anonymous
    • one year ago
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    Lol okay. Now let's use the 1st and 3rd equations so we can get rid of y again.

  75. anonymous
    • one year ago
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    alright

  76. anonymous
    • one year ago
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    |dw:1438713734611:dw|

  77. anonymous
    • one year ago
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    |dw:1438713838855:dw|

  78. anonymous
    • one year ago
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    |dw:1438713940961:dw|

  79. anonymous
    • one year ago
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    Did you get that?

  80. anonymous
    • one year ago
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    noooooooooooooooooo

  81. anonymous
    • one year ago
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    What part did you get lost on?

  82. anonymous
    • one year ago
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    |dw:1438714161470:dw| i dont understand any of it, how do we get to the answers that it wants, why am i so BAD at UNDERSTANDING THINGS

  83. anonymous
    • one year ago
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    Oh we didn't get to finding the answers yet. We needed to find two equations to find x and z. We did those last two steps to get rid of y.

  84. anonymous
    • one year ago
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    why did we multiply that row by 2 though?

  85. anonymous
    • one year ago
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    To get rid of the y. You need to multiply the equation by a certain number to cancel out the y. In that case the y in the 1st equation is 2 while the 3rd equation's y is 1. So if you were to just look at\[2y\]&\[y\]How would you be able to get that y to equal the 2y?

  86. anonymous
    • one year ago
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    multiply it by two but...

  87. anonymous
    • one year ago
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    what happened to the row in the middle

  88. anonymous
    • one year ago
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  89. anonymous
    • one year ago
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    Oh it was cancelled because we need to cancel out variables to find x or z. In order to find what x y and z equals we need to have two variables in two equations in order to solve it. That's why we got rid of y.

  90. anonymous
    • one year ago
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    Did you get that?

  91. anonymous
    • one year ago
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    so...that whole row was canceled out?

  92. anonymous
    • one year ago
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    No just y was cancelled out. We subtracted the two equations to get rid of y and have only x and z left in a new equation.

  93. anonymous
    • one year ago
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    i dont get iiiit

  94. anonymous
    • one year ago
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    Which part?

  95. anonymous
    • one year ago
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    im a kinesthetic learner, i figure things out differently im sorry this must be frustrating.

  96. anonymous
    • one year ago
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    everything

  97. anonymous
    • one year ago
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    |dw:1438715563780:dw|

  98. anonymous
    • one year ago
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    And it's okay.

  99. anonymous
    • one year ago
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    Did you understand this picture?

  100. anonymous
    • one year ago
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    no

  101. anonymous
    • one year ago
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    Do you know where I got the 1st and 3rd equations from?

  102. anonymous
    • one year ago
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    lets solve this whole thing and then break it down because i dont know where we are headed

  103. anonymous
    • one year ago
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    no

  104. anonymous
    • one year ago
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    Oh okay then hold on.

  105. anonymous
    • one year ago
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    its really cool of you to help me

  106. anonymous
    • one year ago
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    thank you

  107. anonymous
    • one year ago
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    \[x+2y-2z=11\]\[5x+4y+7z=81\]\[-4x+y-6z=-37\]----------------------- 1st step:\[x+2y-2z=11\]\[5x+4y+7z=81\]----------------------- \[2(x+2y-2z=11)\]\[5x+4y+7z=81\]----------------------- \[2x+4y-4z=22\]\[5x+4y+7z=81\]----------------------- \[(2x+4y-4z=22)\]minus \[(5x+4y+7z=81)\]equals\[(-3x+0y-11z=-59)\]OR\[-3x-11z=-59\](1st equation with no y's) ----------------------- 2nd step: (Pick another two sets of equations to cancel out y.)\[x+2y-2z=11\]\[-4x+y-6z=-37\]-----------------------\[x+2y-2z=11\]\[2(-4x+y-6z=-37)\]-----------------------\[x+2y-2z=11\]\[-8x+2y-12z=-74\]-----------------------\[x+2y-2z=11\]minus \[-8x+2y-12z=-74\]equals\[9x+0y+10z=85\]OR\[9x+10z=85\](2nd equation with no y's) -----------------------

  108. anonymous
    • one year ago
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    And you're welcome! :) This doesn't give you the answers yet but I'm just going to show you this part to see if you understand it.

  109. anonymous
    • one year ago
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    oooooh

  110. anonymous
    • one year ago
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    You understand it now?

  111. anonymous
    • one year ago
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    yes

  112. anonymous
    • one year ago
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    Okay great! I should have just showed you the steps earlier like that lol. XD

  113. anonymous
    • one year ago
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    yeah omg

  114. anonymous
    • one year ago
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    1st equation with no y's:\[-3x-11z=-59\]2nd equation with no y's:\[9x+10z=85\]Now we need to get rid of x's in this step since that's the easiest to multiply with. We need to get -3x equal to 9x. We have to multiply the 1st equation with no y's by 3.\[3(-3x-11z=-59)\]\[9x+10z=85\]-----------------------\[-9x-33z=-177\]\[9x+10z=85\]-----------------------\[-9x-33z=-177\]plus \[9x+10z=85\]equals\[0x-23z=-92\]OR\[-23z=-92\]----------------------- Now that we have one variable left, we can finally find what z equals. Divide -92 by -23.\[z=\frac{ -92 }{ -23 }\]\[z=4\]----------------------- After finding one variable, you can now substitute the z in the equation with no y's. You can do either the 1st or 2nd equation with no y's. It doesn't matter.\[9x+10z=85\]\[9x+10(4)=85\]\[9x+40=85\]\[9x=85-40\]\[9x=45\]\[x=5\]----------------------- Lastly we can solve for y by using the equations givin to us in the beginning of the question (from the matrices). Either 3 of the equations can work. Make sure to substitute the x and z.\[x+2y-2z=11\]\[(5)+2y-2(4)=11\]\[5+2y-8=11\]\[2y-3=11\]\[2y=11+3\]\[2y=14\]\[y=7\]

  115. anonymous
    • one year ago
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    oh wow

  116. anonymous
    • one year ago
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    Tell me if you understand that. This is the final step and it gives the answer.

  117. anonymous
    • one year ago
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    that is a process

  118. anonymous
    • one year ago
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    Ya it might look really long but don't let that scare you. Once you keep practicing it, it will be really easy! :)

  119. anonymous
    • one year ago
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    omg i actually understand

  120. anonymous
    • one year ago
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    Yay that's great! :D

  121. anonymous
    • one year ago
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    see, im taking a summer course and its online, the section imtaking is all about that so im stuck

  122. anonymous
    • one year ago
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    Oh okay. Well hopefully this was helpful in figuring out how to solve it. :)

  123. anonymous
    • one year ago
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    it was very helpful!

  124. anonymous
    • one year ago
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    Lol I'm glad!

  125. anonymous
    • one year ago
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    i might need help going through another problem because i know there are exceptions

  126. anonymous
    • one year ago
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    |dw:1438718495088:dw|

  127. anonymous
    • one year ago
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    Okay. If it's another problem can you write it in the Ask a question box? That way I can scroll through the steps that belong to that problem?

  128. anonymous
    • one year ago
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    oh sure

  129. anonymous
    • one year ago
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    Okay thanks!

  130. anonymous
    • one year ago
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    so do i get rid of the x?

  131. anonymous
    • one year ago
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    because 4 can go into the 8 or no

  132. anonymous
    • one year ago
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    |dw:1438718635151:dw|

  133. anonymous
    • one year ago
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    Ya you can get rid of the x or y since you can just multiply the 3rd equation with 4 so it can be the same x and y in the 2nd equation.

  134. anonymous
    • one year ago
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    |dw:1438718776581:dw|

  135. anonymous
    • one year ago
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    |dw:1438718875081:dw|