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jagr2713

  • one year ago

x-3y=-9 6x+y=3

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  1. jagr2713
    • one year ago
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    i dont understand this @freckles

  2. steve816
    • one year ago
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    How does a 99 rated mathlete not know how to do system of equations O.o

  3. jagr2713
    • one year ago
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    Well when you haven't done math in a while u forget things.....

  4. IrishBoy123
    • one year ago
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    |dw:1442268783821:dw|

  5. jagr2713
    • one year ago
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    ok i know the part where u add them and stuff but i just need help in the beginning. How do u find what to multiply them with?

  6. IrishBoy123
    • one year ago
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    when they are set out like that, just imagine adding them and knocking out an x or a y so, let's do it again

  7. IrishBoy123
    • one year ago
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    |dw:1442269218613:dw|

  8. jagr2713
    • one year ago
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    wait why did u use 1 and 2

  9. IrishBoy123
    • one year ago
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    if you meant, the (1) and (2) in brackets to identify the equations, and just a habit is that what you were referring to?

  10. jagr2713
    • one year ago
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    Because my teacher taught me to multiply them by a number and then add then and cancel the x or y u used etc

  11. IrishBoy123
    • one year ago
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    loads of ways to do these eg |dw:1442269699628:dw|

  12. IrishBoy123
    • one year ago
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    can i ask , do you work on the OS website??

  13. jagr2713
    • one year ago
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    whats the easiest way? can u do in equation form cause the drawing is losing me

  14. jagr2713
    • one year ago
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    do i work on the website?

  15. IrishBoy123
    • one year ago
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    yes

  16. jagr2713
    • one year ago
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    what does that mean?

  17. IrishBoy123
    • one year ago
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    are you part of the OS team?

  18. jagr2713
    • one year ago
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    Oh yes...

  19. IrishBoy123
    • one year ago
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    easiest way is substitution, IMHO. as in, it's pure method. so, from the first equation, which is \(x - 3y = -9\), you can say \(x = 3y - 9\) then in the second, which is \(6x + y = 3\), you can say \(6(3y - 9) +y = 3\) so \(18y - 54 + y = 3\) or \(19y = 57\), \(y = 3\)

  20. IrishBoy123
    • one year ago
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    so you code this site?!

  21. jagr2713
    • one year ago
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    OOOHOHO you have to change the x-3y=*9 in x=3y-9 and lets stay relevant :D u can message me the other questions

  22. IrishBoy123
    • one year ago
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    OK!!

  23. jagr2713
    • one year ago
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    But what if its x+3y=-9 does 3y become positive cause u move it?

  24. IrishBoy123
    • one year ago
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    if x + 3y = -9 then x = -3y - 9

  25. IrishBoy123
    • one year ago
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    because x+3y-3y = -9-3y

  26. jagr2713
    • one year ago
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    so i got 6(3y-9)+y=3 18y-54+y=3 +y +y 19y-54=3 +54 +54 19y=57 y=3

  27. jagr2713
    • one year ago
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    so now u plug it into the x too get x right? x-3(3)=-9 x+9=-9 -9 -9 x=-18 so we got (-18,3)

  28. jagr2713
    • one year ago
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    ??

  29. IrishBoy123
    • one year ago
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    if you know that y = 3, plug that into one of the original equations to find x

  30. IrishBoy123
    • one year ago
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    eg, if x - 3y = -9 what is x if y = 3

  31. IrishBoy123
    • one year ago
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    or if 6x + y = 3 what is x if y = 3??

  32. jagr2713
    • one year ago
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    you plug it into the x-3y=-9 right?

  33. IrishBoy123
    • one year ago
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    into either, you want to know when they meet

  34. jagr2713
    • one year ago
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    it says -18,3 is wrong

  35. jagr2713
    • one year ago
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    wait don't i need another point?

  36. IrishBoy123
    • one year ago
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    i thought we had agreed 0, 3 ?!?!

  37. IrishBoy123
    • one year ago
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    as in x = 0

  38. IrishBoy123
    • one year ago
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    and y = 3

  39. jagr2713
    • one year ago
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    OOHOHOHO i did it wrong ok so i need another set of points to connect it.

  40. IrishBoy123
    • one year ago
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    indeed you might! gtg

  41. jagr2713
    • one year ago
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    So how do i get it?

  42. jagr2713
    • one year ago
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    @freckles

  43. IrishBoy123
    • one year ago
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    really sorry @jagr2713 i gotta go sleep i am in europe, i guess you are somewhere else my condolensces!!!!

  44. jagr2713
    • one year ago
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    Its ok

  45. jdoe0001
    • one year ago
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    are you meant to solve by either substitution or elimination?

  46. jagr2713
    • one year ago
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    Graphing them

  47. jagr2713
    • one year ago
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    i got the answer 0,3 just need another point

  48. jdoe0001
    • one year ago
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    well... where the graph of two equations meet, is where the answer is :) hm if 0,3 is the answer.... isn't that what you needed? :)

  49. jagr2713
    • one year ago
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    Yea but thats only one point need 2 for a line

  50. jdoe0001
    • one year ago
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    hmmm one sec

  51. jagr2713
    • one year ago
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    ok :D

  52. jagr2713
    • one year ago
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    here use this i did another question x-3y=-3 3x+y=1 i got (0,1)

  53. jdoe0001
    • one year ago
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    ..is that another one?

  54. jagr2713
    • one year ago
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    Yea i just change questions i got 0,1 but need another point

  55. jdoe0001
    • one year ago
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    hmmm thought you were supposed to just graph them?

  56. jagr2713
    • one year ago
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    Yea it say solve the following system of equation by graphing them

  57. jdoe0001
    • one year ago
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    well... I'd just graph them then :) lemme do a quick table

  58. jagr2713
    • one year ago
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    ok thanks

  59. jagr2713
    • one year ago
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    wait do we just do the x|y thing and put 0 for x and ory?

  60. jagr2713
    • one year ago
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    |dw:1442273690049:dw| like that?

  61. jdoe0001
    • one year ago
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    yeap to make a line, all you need is 2 points so just pick two random "x" to get two "y"s

  62. jagr2713
    • one year ago
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    OHH so |dw:1442273743049:dw|

  63. jagr2713
    • one year ago
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    other is 0,3 2,0

  64. jdoe0001
    • one year ago
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    hmm yes \(\begin{array}{llll} x-3y=-3\implies &\cfrac{x+3}{3}={\color{brown}{ y}} \\ \quad \\ 3x+y=1\implies &{\color{brown}{ y}}=1-3x \end{array} \\ \quad \\ \qquad \qquad \begin{array}{ccllll} x&y \\\hline\\ &\frac{x+3}{3}\\ \\\hline\\ &\frac{x+3}{3}\\ 0&1\\ -3&0 \end{array}\qquad \begin{array}{ccllll} x&y \\\hline\\ &1-3x \\\hline\\ 0&1\\ 4&-11 \end{array}\)

  65. jagr2713
    • one year ago
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    where did u get they -3 from?

  66. jdoe0001
    • one year ago
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    x-3y=-3 <--- 3x+y=1

  67. jagr2713
    • one year ago
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    is this correct (0,3) (-9,0) and (0,3) (2,0) ?

  68. jdoe0001
    • one year ago
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    hmmm 0, 3 is not, -9,0 isn't either

  69. jdoe0001
    • one year ago
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    x =0 x -3y=-3 0-3y=-3 y = (-3)/(-3) y =1

  70. jagr2713
    • one year ago
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    x-3y=-9 -3y=-9 y=1/3 x-3y=-9 x=-9

  71. jdoe0001
    • one year ago
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    x=-9 (-9)-3y=-3 -3y=-3+9 -3y=+6 y=(6)/(-3) y=-2

  72. jagr2713
    • one year ago
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    wait this is confusing me dont u put 0 for them?

  73. jdoe0001
    • one year ago
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    wel... you said it was x-3y=-3 3x+y=1 but now you seem to have changed it some

  74. jagr2713
    • one year ago
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    OHOHOHO I've been doing the wrong problem... :D KML

  75. jagr2713
    • one year ago
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    I WAS LOOKING AT THE QUESTION LOL

  76. jagr2713
    • one year ago
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    i got (0,1) (-3,0) and (0,1) (3,0)

  77. jdoe0001
    • one year ago
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    but all you need to graph a line is two points \(\begin{array}{llll} x-3y=-3\implies \cfrac{x+9}{3}={\color{brown}{ y}}\\ 6x+y=3\implies {\color{brown}{ y}}=3-6x \end{array}\) http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIoeCs5KS8zIiwiY29sb3IiOiIjM0IxRENGIn0seyJ0eXBlIjowLCJlcSI6IjMtNngiLCJjb2xvciI6IiNENDExMTEifSx7InR5cGUiOjEwMDAsIndpbmRvdyI6WyItNi41NCIsIjYuNDYiLCItMi40OCIsIjUuNTIiXX1d

  78. jagr2713
    • one year ago
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    D: it says my answer is wrong... i did the x|y thing

  79. jdoe0001
    • one year ago
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    check yours PEMDAS I'd think

  80. freckles
    • one year ago
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    I'm confused... Are you saying you got 3 different solutions for x-3y=-3 3x+y=1 ? You should only get one solution, infinitely many solutions, or no solutions for system of linear equations.

  81. jagr2713
    • one year ago
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    OHH my friend told me to use y=mx+b

  82. freckles
    • one year ago
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    why? aren't you trying to solve the system x-3y=-3 3x+y=1 ? or is that not the question?

  83. jagr2713
    • one year ago
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    yes by graphing it thou but i did it and getting it wrong D:

  84. freckles
    • one year ago
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    you should just get a point for the solution of this system: you will see that when graphing these two lines that they will share a common point that point is the solution x-3y=-3 -3y=-x-3 y=1/3*x+1 --- 3x+y=1 y=-3x+1 so graphing both y=1/3*x+1 and -3x+1 we have something that looks like this: |dw:1442275696567:dw| the solution is given at the point where both lines touch each other (cross each other; intersect each other)

  85. freckles
    • one year ago
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    |dw:1442275792808:dw| so this point I darkened is the solution because is the point that two lines have in common

  86. jagr2713
    • one year ago
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    wait can u try thing one x-2y=-2 4x+y=1

  87. freckles
    • one year ago
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    were you trying to write in y=mx+b form to graph the lines to find the point ?

  88. jagr2713
    • one year ago
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    yea i was using that

  89. freckles
    • one year ago
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    ok well try writing x-2y=-2 in that form

  90. jagr2713
    • one year ago
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    2y=x+2 right?

  91. freckles
    • one year ago
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    so you subtracted x on both sides and multiplied -1 on both sides? x-2y=-2 subtracted x on both sides -2y=-x-2 multiplied both sides by -1 2y=x+2 this is correct so far then just divide both sides by 2

  92. jagr2713
    • one year ago
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    yea i got y=1/2x+1

  93. jagr2713
    • one year ago
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    then u graph it by going to (0,1) then rise over run

  94. freckles
    • one year ago
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    that sounds great so y=1/2x+1 says the y-intercept is 1 and the slope is 1/2 |dw:1442276083294:dw| right so you get something like this now we also need to graph 4x+y=1 on that same graph

  95. freckles
    • one year ago
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    4x+y=1 is equivalent to y=-4x+1 can you try graph that we should actually already see the solution even before graphing

  96. jagr2713
    • one year ago
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    yea we got 0,1 then rise over run -4/1

  97. freckles
    • one year ago
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    so before graphing you should see that (0,1) is a point on both graphs where both graphs differ in slope therefore (0,1) is the solution

  98. jagr2713
    • one year ago
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    yea 0,1 is the slope so we go down 4 and move 1 to the left right?

  99. freckles
    • one year ago
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    (0,1) is the y-intercept the slope is -4/1 so we move down from the y-intercept 4 units and to the right once

  100. jagr2713
    • one year ago
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    Oh right 1 sorry my friend told me left....

  101. jagr2713
    • one year ago
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    OMG your the best :D

  102. jagr2713
    • one year ago
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    can u help with one more but this one is substitution

  103. freckles
    • one year ago
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    sure

  104. freckles
    • one year ago
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    one more then I have to roll my croissant dough

  105. jagr2713
    • one year ago
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    ok :D 4x+y=24 y=2x

  106. freckles
    • one year ago
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    replace first y with (2x) solve 4x+2x=24 for x

  107. jagr2713
    • one year ago
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    don't you reverse it like y=4x+24 and plug into x y=2(4x+24)

  108. freckles
    • one year ago
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    the second equation says y=2x so you insert that y here: 4x+y=24 to get 4x+2x=24 anyways you can't replace x with 4x+24 because we aren't given x=4x+24 unless you already know that x=-8 which x doesn't -8 and also 4x+y=24 is not equivalent to y=4x+24

  109. jagr2713
    • one year ago
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    wait how did u get 4x+2x=24

  110. freckles
    • one year ago
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    \[\color{red}{y=2x} \\ 4x+\color{red}{y}=24 \\ \\ 4x+\color{red}{2x}=24\]

  111. freckles
    • one year ago
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    I replaced y with 2x because one of the equations said y=2x

  112. jagr2713
    • one year ago
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    Wait i understand igtg thanks for the help

  113. freckles
    • one year ago
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    k peace

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