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Inopeki

In linear equations, how do i know if i should subtract from the right side or the left side?

  • 2 years ago
  • 2 years ago

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  1. karatechopper
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    example..

    • 2 years ago
  2. meverett04
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    I like my coefficient for x to remain positive; therefore, if I have 2x + 10 = 12x I subtract 2x from the left side and if I have 12x + 10 = 2x I subtract 2x from the right side

    • 2 years ago
  3. Inopeki
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    3x + 7 = 10x +5 7 = 7x +5 - ^here? ^ or here?

    • 2 years ago
  4. TuringTest
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    it really doesn't matter so long as you keep the goal in mind: get x by itself pick one and we can continue from there

    • 2 years ago
  5. Inopeki
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    And how much do i subtract? the whole number or so it remains +/-0?

    • 2 years ago
  6. Inopeki
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    Ok

    • 2 years ago
  7. Inopeki
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    so i got 0=7x-2

    • 2 years ago
  8. Inopeki
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    now i need to divide -2 by 7?

    • 2 years ago
  9. TuringTest
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    no, we need the 2 on the other side of the equation that's what I mean about getting x by itself

    • 2 years ago
  10. Inopeki
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    Ok so i dont do 0=7x-2?

    • 2 years ago
  11. TuringTest
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    if we divide by 7 now we get\[0=x-\frac{2}{7}\]which is okay, we can still solve it from here; but it's a good policy to leave fractions for the end if possible. so instead add 2 to both sides to make x a little bit more alone.

    • 2 years ago
  12. Inopeki
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    I dont get it, i get like 0.29..

    • 2 years ago
  13. TuringTest
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    leave the answer in fraction form

    • 2 years ago
  14. Inopeki
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    Oh i see, so i get 2=7x?

    • 2 years ago
  15. TuringTest
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    right, now we divide by 7 to isolate x what is x=? (leave as a fraction)

    • 2 years ago
  16. Inopeki
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    I dont know.. 2/7?

    • 2 years ago
  17. TuringTest
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    you do know! lol shall we do another?

    • 2 years ago
  18. phi
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    If you want more info, watch this and the next 2 videos http://www.khanacademy.org/video/simple-equations?playlist=Algebra

    • 2 years ago
  19. TuringTest
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    that's what he's working from phi

    • 2 years ago
  20. Inopeki
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    What?

    • 2 years ago
  21. TuringTest
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    Another?

    • 2 years ago
  22. Inopeki
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    Sure

    • 2 years ago
  23. Inopeki
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    Well ill be damned.. it was right!

    • 2 years ago
  24. TuringTest
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    got one you want to post? or should I make one up?

    • 2 years ago
  25. Inopeki
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    wait, ill try the next one. 6x+3=2x+7

    • 2 years ago
  26. TuringTest
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    just go ahead and see how far you get

    • 2 years ago
  27. Inopeki
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    So i subtract 2x from both sides, getting 4x+3=7

    • 2 years ago
  28. TuringTest
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    good, then?

    • 2 years ago
  29. Inopeki
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    Then i subtract 3 from both sides, getting 4x=4

    • 2 years ago
  30. TuringTest
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    good, then?

    • 2 years ago
  31. Inopeki
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    And then i divide 4 by 4 and get 1 as the final answer :D

    • 2 years ago
  32. karatechopper
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    :D

    • 2 years ago
  33. karatechopper
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    well taught turing:)

    • 2 years ago
  34. Inopeki
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    TT, you are an awesome teacher!

    • 2 years ago
  35. TuringTest
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    tadah! nice see how you didn't need to ask me which was better, subtract 7 or 3? you knew by looking that 3 was the smart way to go :) practice practice... thanks for the compliment btw

    • 2 years ago
  36. Inopeki
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    So where do i go on from here?

    • 2 years ago
  37. TuringTest
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    have you done ALL the exercises from this section already?

    • 2 years ago
  38. Inopeki
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    Soon. you mean until i can choose another section right?

    • 2 years ago
  39. TuringTest
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    Yes. It is imperative that you do every problem. This needs to become second nature for you when you can look at 5x-3=2x+12 and tell me the answer very quickly, with almost no work shown, then you can continue

    • 2 years ago
  40. Inopeki
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    3/4?

    • 2 years ago
  41. Inopeki
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    actually 1/3

    • 2 years ago
  42. TuringTest
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    nope, better keep working on it. practice practice... don't try to get ahead of yourself, if you need to write it and go slowly, do it. Don't try to shortcut past that.

    • 2 years ago
  43. Inopeki
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    alright. but i couldnt figure the last one out, it was 6x+6=0

    • 2 years ago
  44. Inopeki
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    I didnt know what to do there. was it 1 or 0? maybe -6?

    • 2 years ago
  45. TuringTest
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    just try the problem and show me your steps you don't need to tell me what each step is, btw, I can see by looking

    • 2 years ago
  46. Inopeki
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    it was 9x+9=3x+3 6x+9=3 6x+6=0 6x= -6?

    • 2 years ago
  47. TuringTest
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    so now what is the last step to get x all alone?

    • 2 years ago
  48. karatechopper
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    tht was correct now show us your next and final step

    • 2 years ago
  49. Inopeki
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    6/-6?

    • 2 years ago
  50. karatechopper
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    perfect

    • 2 years ago
  51. karatechopper
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    now simplify that

    • 2 years ago
  52. Inopeki
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    -1?

    • 2 years ago
  53. karatechopper
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    correct

    • 2 years ago
  54. Inopeki
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    Yes!

    • 2 years ago
  55. TuringTest
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    yes, very good what was the last step? dividing by...?

    • 2 years ago
  56. karatechopper
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    that wasn't too bad now was it!

    • 2 years ago
  57. Inopeki
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    The number on the right?

    • 2 years ago
  58. Inopeki
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    dividing the quantity of x by the number on the right side of =?

    • 2 years ago
  59. Inopeki
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    Other*

    • 2 years ago
  60. TuringTest
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    no, that is what I was afraid of... though you got the right answer, the -6 should have stayed on top. 6x=-6 what do we divide by to isolate x?

    • 2 years ago
  61. Inopeki
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    -6 with 6?

    • 2 years ago
  62. TuringTest
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    it's just one answer: divide by 6

    • 2 years ago
  63. TuringTest
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    let's try something a little different... solve xy+7=10 for x

    • 2 years ago
  64. Inopeki
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    Ok

    • 2 years ago
  65. TuringTest
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    Note that above for 6x=-6 we divided both sides by the number next to the x (i.e. 6), that is what I wanted you to see. We do not divide by the number on the right. You will see why in this next problem

    • 2 years ago
  66. Inopeki
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    1.5 or 2 for x

    • 2 years ago
  67. Inopeki
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    Ok..

    • 2 years ago
  68. TuringTest
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    so now xy+7=10 you got a number answer? what happened to y? please show steps.

    • 2 years ago
  69. karatechopper
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    @turing does he know what he is solving for?

    • 2 years ago
  70. TuringTest
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    solve for x thanks karate

    • 2 years ago
  71. karatechopper
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    anytime turing:)

    • 2 years ago
  72. Inopeki
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    I thought that x times y would be the answer. since 1.5x2 is 3 (which was needed to make 10) i thought that x would be 1.5 or 2.(the other one being y)

    • 2 years ago
  73. TuringTest
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    you are over-thinking Inopeki just take each step one-by-one what is the first step?

    • 2 years ago
  74. Inopeki
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    Determening what i need to make to complete 10

    • 2 years ago
  75. TuringTest
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    no the first step is the same as the others xy+7=10 subtract 7 from both sides

    • 2 years ago
  76. TuringTest
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    what do we have after?

    • 2 years ago
  77. Inopeki
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    Ah.. xy=3

    • 2 years ago
  78. TuringTest
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    good! now again, we want x by itself. it looks like x is being multiplied by y, so how can we get x by itself?

    • 2 years ago
  79. Inopeki
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    Removing y from the equation?

    • 2 years ago
  80. TuringTest
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    y cannot leave the equation entirely, we can only shift it to one side. I mentioned that x was being multiplied by y. Do you know the inverse operation of multiplication?

    • 2 years ago
  81. Inopeki
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    division?

    • 2 years ago
  82. TuringTest
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    right! and in this case what do we need to divide by to get x by itself?

    • 2 years ago
  83. Inopeki
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    y with 3?

    • 2 years ago
  84. TuringTest
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    close, but what do you mean "divide by y with 3"? that makes no sense again it can only be one answer divide by...?

    • 2 years ago
  85. TuringTest
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    just to make it visible again: xy=3 what do we divide by on both sides (one answer to this question)

    • 2 years ago
  86. Inopeki
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    Ah, divide x and 3 by y?

    • 2 years ago
  87. TuringTest
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    YES!!! and what do we have after? write the expression.

    • 2 years ago
  88. Inopeki
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    How can i do that when i dont even know what y is?

    • 2 years ago
  89. TuringTest
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    Good question. write the answer first and I will tell you what it means after.

    • 2 years ago
  90. Inopeki
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    You mean x/y=3/y?

    • 2 years ago
  91. TuringTest
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    close, but on the left x should be by itself, that's why we divided by y right? so it should be...

    • 2 years ago
  92. Inopeki
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    I GET IT NOW! because the left side was x times y we needed to isolate x so we divided x with y to reverse the effect. now it is x=3/y

    • 2 years ago
  93. TuringTest
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    YES!!! very good! but what does it mean...? as you said y can be any number (except zero, but we'll discuss that later) so we have x=3/y that means we have infinite solutions to this problem. To find what they are plug in various numbers for y. So what is x if y=1 y=3 y=10 ???

    • 2 years ago
  94. Inopeki
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    1.5?

    • 2 years ago
  95. TuringTest
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    no, its different for each one. Remember that I said there were infinite solutions. just plug in the numbers into the equation there is no one answer for x=3/y so plug in a number for y like y=1 what is x?

    • 2 years ago
  96. Inopeki
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    3

    • 2 years ago
  97. TuringTest
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    yes! what about when y=3?

    • 2 years ago
  98. Inopeki
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    1

    • 2 years ago
  99. TuringTest
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    Excellent :) and y=10?

    • 2 years ago
  100. Inopeki
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    0.3333333?

    • 2 years ago
  101. TuringTest
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    better to leave it as a fraction, but yes

    • 2 years ago
  102. Inopeki
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    YES! I get it now!

    • 2 years ago
  103. TuringTest
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    Great! so what is going on here? it looks like we can put any value for y (except zero) and we get a number for x. THIS IS YOUR FIRST FUNCTION I told you earlier that a function takes a number as an input (in this case what we put for y is our input) and out pops another number. So x=3/y is a formula for x as a function of y. denoted x=f(y)=3/y here is a graph of your function http://www.wolframalpha.com/input/?i=f%28y%29%3D3%2Fy this line denotes all the points that are solutions. notice that some of these points (x,y) are (1,3) (3,1) (10,3/10) etc. just like you found :)

    • 2 years ago
  104. Inopeki
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    Whoa, whoa, whoa. You just made this 10x more complicated.

    • 2 years ago
  105. TuringTest
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    look all the deal is is that you put in a number for y, you get a number back for x that's what a function is

    • 2 years ago
  106. Inopeki
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    Oh! I just made a function?

    • 2 years ago
  107. TuringTest
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    Yes. Well actually it was always a function, buit you made it workable. in this case we would write that x is a function of y. f(y)=3/y

    • 2 years ago
  108. Inopeki
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    So f(x) means that you want to pop something for y?

    • 2 years ago
  109. TuringTest
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    exactly! if you want to find f(x) then solve\[xy+7=10\]for x instead of y

    • 2 years ago
  110. TuringTest
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    sorry, I meant y instead of x above

    • 2 years ago
  111. Inopeki
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    So thats what ill be using in quadratic equalities?

    • 2 years ago
  112. TuringTest
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    it's one thing you will need in order to understand them. in general though, you should not be thrown by seeing f(x)=5x+7 that is just the same situation we had. they are saying that f(x) depends on what number you put in for x. you may see 'find when f(x)=10 for f(x)=5x+5' in which case you just use the equation, and put the value of f(x) on the other side of the = sign: 10=5x+5 5=5x x=1 we can do that for any value of f(x)

    • 2 years ago
  113. Inopeki
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    Shouldnt x be 2 there?

    • 2 years ago
  114. TuringTest
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    where?

    • 2 years ago
  115. TuringTest
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    the problem? no follow it closely

    • 2 years ago
  116. Inopeki
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    Oh nevermind

    • 2 years ago
  117. Inopeki
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    Can you give me one so i can try?

    • 2 years ago
  118. TuringTest
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    sure, solve 5t+2r=3r-4 for r

    • 2 years ago
  119. Inopeki
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    I see what you are doing, you are trying to flutter up my train of thoughts by replacing x and y with t and r.. hah, itll never work.

    • 2 years ago
  120. Inopeki
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    flutter?

    • 2 years ago
  121. TuringTest
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    good! better get used to it, all kinds of funny symbols in physics :P

    • 2 years ago
  122. TuringTest
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    flutter is how they translate the f-word lol

    • 2 years ago
  123. Inopeki
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    5t+2r=3r-4 5t=r-4 Am i right so far?

    • 2 years ago
  124. TuringTest
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    yes, now what?

    • 2 years ago
  125. Inopeki
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    Well i need you to tell me what t is. Ill just assume it is 1. 5t=9(r)-4?

    • 2 years ago
  126. TuringTest
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    no you don't. we are going to get a function out of this because we have two variable and only one eqn. so how do you isolate r?

    • 2 years ago
  127. TuringTest
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    5t=r-4

    • 2 years ago
  128. Inopeki
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    5t-4=r?

    • 2 years ago
  129. TuringTest
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    perfect! so do you think you can tell me 'what is a function of what' in this case?

    • 2 years ago
  130. Inopeki
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    f(t)?

    • 2 years ago
  131. TuringTest
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    yes, that's right! sorry btw, you made a mistake earlier: 5t=r-4 ADD 4 to both sides: 5t+4=r but you got the fact that this is a function of t, very good. which variable is a function of t?

    • 2 years ago
  132. Inopeki
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    r

    • 2 years ago
  133. Inopeki
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    It must be because 4 is an insteger, right?

    • 2 years ago
  134. TuringTest
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    perfect :) (the integer thing makes no difference. If it was\[r=5t+\sqrt2\]that would be a function as well. we can write our function as r=f(t)=5t+4 or we can also write r(t)=5t+4 both mean that r is a function of t. Same exact thing. Now I'm going to use the notation r(t)=5t+4 and I am asking you, what is f(0)=? f(1)=? f(-1)=?

    • 2 years ago
  135. TuringTest
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    sorry, I meant r(t)=5t+4 what is r(0)=? r(1)=? r(-1)=?

    • 2 years ago
  136. Inopeki
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    I thought we determined that r(0) cant be a variable?

    • 2 years ago
  137. TuringTest
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    that was only when the 'independent variable' (that's what we call the t in this case) was in the denominator, because any number divided by zero is undefined (infinity basically) here we don't have that problem, so just plug in the numbers and see what you get.

    • 2 years ago
  138. TuringTest
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    r is called the 'dependent variable' btw, because it 'depends' on what we put for t.

    • 2 years ago
  139. Inopeki
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    I see

    • 2 years ago
  140. TuringTest
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    so find r(0) r(1) and r(-1) for r=5t+4

    • 2 years ago
  141. Inopeki
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    r(t)=5t+4 what is r(0)=-1/4 r(1)=1/9 r(-1)=-5

    • 2 years ago
  142. Inopeki
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    Thats probably really wrong..

    • 2 years ago
  143. TuringTest
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    no, show your work... r(0)=5(0)+4=4 do the same for the others

    • 2 years ago
  144. Inopeki
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    OH! so (this) is what equals t?

    • 2 years ago
  145. TuringTest
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    right!

    • 2 years ago
  146. Inopeki
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    r(t)=5t+4 what is r(0)=5(0)+4=4 r(1)=5(1)+4 So it equals 9! r(-1)=5(-1)+4=-1 ((-5)+4=-1)

    • 2 years ago
  147. Inopeki
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    That should do it.

    • 2 years ago
  148. TuringTest
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    excellent!

    • 2 years ago
  149. Inopeki
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    YEEEEEEEEES!!!!!!!!

    • 2 years ago
  150. TuringTest
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    great, so now you have dealt with two functions today. do you know how to plot points on a graph?

    • 2 years ago
  151. Inopeki
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    |dw:1325547966414:dw|

    • 2 years ago
  152. TuringTest
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    yes! now do you know what r(0)=4 means as a point?

    • 2 years ago
  153. Inopeki
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    As a point?

    • 2 years ago
  154. TuringTest
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    yes, it is the point (0,4) we can call the up-down direction on our graph r, and the horizontal direction t. this way we get a point out of every value of t we put in. the points (t,r) that we found are then r(0)=4-->(0,4) r(1)=9--->(1,9) r(-1)=-1-->(-1,-1) ^^^don't continue until you understand all this, or at least have tried.

    • 2 years ago
  155. Inopeki
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    Ok, give me an example and ill try it. like r(2)=4---->2,4?

    • 2 years ago
  156. TuringTest
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    here is our graph marked from 0 to 4 in all directions:|dw:1325548510639:dw|can you plot two of the points? If you want to try some more first that's even better: -find the point (actually it's called an 'ordered pair') that corresponds to r(-2) for r(t)=5t+4

    • 2 years ago
  157. TuringTest
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    Actually, just do the question for now: -find the ordered pair that corresponds to r(-2) for r(t)=5t+4

    • 2 years ago
  158. Inopeki
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    r(-2)=5(-2)+4=-6

    • 2 years ago
  159. TuringTest
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    good, so what is the ordered pair (t,r) ?

    • 2 years ago
  160. Inopeki
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    -2,5?

    • 2 years ago
  161. TuringTest
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    -2 is right, but where did 5 come from? that's not the value of r(-2)... and please put it in parentheses, that is how ordered pairs are written.

    • 2 years ago
  162. Inopeki
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    I see, (-2,-6)?

    • 2 years ago
  163. TuringTest
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    perfect :D wanna try to graph it yet, or find more ordered pairs?

    • 2 years ago
  164. Inopeki
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    One more ordered pair would be good.

    • 2 years ago
  165. TuringTest
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    ok, find the ordered pairs that correspond to r(-3) r(3) and r(5) for r=5x+4

    • 2 years ago
  166. TuringTest
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    sorry, you said one more you can only do one if you want.

    • 2 years ago
  167. Inopeki
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    Thats alright, ill do all of them. r(-3)=5(-3)+4=-11 The corresponding pair is (-3,-11) r(3)=5(3)+4=19 The corresponding pair is (3,19) r(5)=5(5)+4=29 The corresponding pair is (5,29)

    • 2 years ago
  168. TuringTest
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    wow!!! you are a very fast learner Inopeki, you'll be doing calculus before you know it ;) so shall we try to graph it now?

    • 2 years ago
  169. Inopeki
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    YES! Thanks, the teacher is just as important though! Sure.

    • 2 years ago
  170. TuringTest
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    thanks for that :D so let's lay out some points that might fit on our graph: r(-1)=-1->(-1,-1) r(0)=4-->(0,4) can you plot those points onto this graph?|dw:1325549886229:dw|(use 'reply with drawing')

    • 2 years ago
  171. Inopeki
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    |dw:1325549850006:dw|

    • 2 years ago
  172. TuringTest
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    just put dots at their coordinates:|dw:1325550068136:dw|now because this graph is linear, what will we connect those two points by?

    • 2 years ago
  173. Inopeki
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    |dw:1325550114671:dw|

    • 2 years ago
  174. TuringTest
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    good, but the line goes on to infinity in both directions:|dw:1325550191308:dw|the arrows mean it keeps going. Good job though you have the idea. Here is a plot of our graph on two different scales: http://www.wolframalpha.com/input/?i=r%28t%29%3D5t%2B4 it doesn't look exactly like ours, but that is only because they made the increments of r larger because the graph is so steep. They wanted to fit more on there.

    • 2 years ago
  175. Inopeki
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    So now what?

    • 2 years ago
  176. TuringTest
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    you can see on the graph I linked you to that the point (1,9) is on there, just as it should be. It also looks like the graph shows that the point (-0.8,0) should be as well. Why don't you see if you can show that the point (-0.8,0) is on our graph. Any idea how to do that?

    • 2 years ago
  177. Inopeki
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    |dw:1325550664006:dw|

    • 2 years ago
  178. Inopeki
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    NO! WAIT!

    • 2 years ago
  179. TuringTest
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    no, we want to do it mathematically, not graphically. Questions should only be answered graphically if they specifically request it. got a better way to show that (-0.8,0) is a point?

    • 2 years ago
  180. Inopeki
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    Oh...

    • 2 years ago
  181. TuringTest
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    well that is the graphic solution, and it looks like you are right. But we want to PROVE that you are right. so how can we do that? think about this: if (-0.8,0) is a point on the graph of r=5t+4 then what should r(-0.8)=?

    • 2 years ago
  182. Inopeki
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    r(0.8)=t(0.8)? r(0.8)=t(0)+0.8?

    • 2 years ago
  183. Inopeki
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    oh wait, ill take that bacl

    • 2 years ago
  184. TuringTest
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    that's a point in the middle, not a comma, and it's negative: t=-0.8 go again!

    • 2 years ago
  185. Inopeki
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    Should i follow r=5t+4?

    • 2 years ago
  186. TuringTest
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    yep :)

    • 2 years ago
  187. Inopeki
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    Ohhh.. r(-0.8)=(-0.8)+4=3.2?

    • 2 years ago
  188. TuringTest
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    what happened to the 5? r=5t+4

    • 2 years ago
  189. Inopeki
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    Oh right, i knew i forgot something. r(-0.8)=5(-0.8)+4=0!

    • 2 years ago
  190. TuringTest
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    which makes sense right? the ordered pair corresponding to that is (-0.8,0) which is what we wanted to prove, that this point is on the graph. Awesome job!

    • 2 years ago
  191. Inopeki
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    Thanks!

    • 2 years ago
  192. TuringTest
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    I'm sure you've learned a lot today, please continue on your own working off khan academy. I do have to eat and stuff, so I'll catch you later. Again, super job! Keep practicing :D

    • 2 years ago
  193. Inopeki
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    Alright :( Well thanks for all your help! It was really fun! Where do i start at khan now?

    • 2 years ago
  194. TuringTest
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    did you finish ALL of that other lecture and problems? if you really finished it all, then you should do the next-to-last one that James gave you, then the first one he gave you. Most importantly, keep working on moving the symbols around. If you run out of exercises here are some to keep you busy. solve: 13+s/p=2 for p 5qr+6=qr/2 for q z/t+z=4t for z You shouldn't be able to do those so easily, as they involve factoring and such. If they give you too much trouble continue with the links James sent you. See you later tonight maybe, otherwise perhaps tomorrow. Good luck!

    • 2 years ago
  195. TuringTest
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    Here it is a little more clear. Again, if you can't do it it's okay. solve\[13+\frac{s}{p}=2\]for p\[5qr+6=\frac{qr}{2}\]for q\[\frac{z}{t}+z=4t\]for z

    • 2 years ago
  196. Inopeki
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    The 3 he gave me were linear equations 3, intro to the quadratic equation and simple equations, im soon finished with linear and then im gonna start with quadratic

    • 2 years ago
  197. Inopeki
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    Wait, on the first one i need to do this: 13= s/p-2 right?

    • 2 years ago
  198. TuringTest
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    actually the first step on the first one is 1)multiply both sides by p the last one above will have a 'squared' term in it, so you may need the lectures first, I don't know. you're on the way to being a physicist, step-by-step again, good luck!

    • 2 years ago
  199. Inopeki
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    Thanks :)

    • 2 years ago
  200. Inopeki
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    p(13)=2(13)+2=28? What is s? Ill say 2. \[28/13\approx2.15\]

    • 2 years ago
  201. Inopeki
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    5qr+6=qr2 4qr+6=qr I can now determine that q*r is negative because otherwise 4qr+6=qr The possibilities of what q and r are endless, what now?

    • 2 years ago
  202. Inopeki
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    TT?

    • 2 years ago
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