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Inopeki

  • 2 years ago

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

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  1. karatechopper
    • 2 years ago
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    example..

  2. meverett04
    • 2 years ago
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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

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

  4. TuringTest
    • 2 years ago
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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

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

  6. Inopeki
    • 2 years ago
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    Ok

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

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

  9. TuringTest
    • 2 years ago
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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

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

  11. TuringTest
    • 2 years ago
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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.

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

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

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

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

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

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

  18. phi
    • 2 years ago
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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

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

  20. Inopeki
    • 2 years ago
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    What?

  21. TuringTest
    • 2 years ago
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    Another?

  22. Inopeki
    • 2 years ago
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    Sure

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

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

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

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

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

  28. TuringTest
    • 2 years ago
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    good, then?

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

  30. TuringTest
    • 2 years ago
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    good, then?

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

  32. karatechopper
    • 2 years ago
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    :D

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

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

  35. TuringTest
    • 2 years ago
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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

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

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

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

  39. TuringTest
    • 2 years ago
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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

  40. Inopeki
    • 2 years ago
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    3/4?

  41. Inopeki
    • 2 years ago
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    actually 1/3

  42. TuringTest
    • 2 years ago
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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.

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

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

  45. TuringTest
    • 2 years ago
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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

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

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

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

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

  50. karatechopper
    • 2 years ago
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    perfect

  51. karatechopper
    • 2 years ago
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    now simplify that

  52. Inopeki
    • 2 years ago
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    -1?

  53. karatechopper
    • 2 years ago
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    correct

  54. Inopeki
    • 2 years ago
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    Yes!

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

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

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

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

  59. Inopeki
    • 2 years ago
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    Other*

  60. TuringTest
    • 2 years ago
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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?

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

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

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

  64. Inopeki
    • 2 years ago
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    Ok

  65. TuringTest
    • 2 years ago
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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

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

  67. Inopeki
    • 2 years ago
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    Ok..

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

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

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

  71. karatechopper
    • 2 years ago
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    anytime turing:)

  72. Inopeki
    • 2 years ago
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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)

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

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

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

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

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

  78. TuringTest
    • 2 years ago
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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?

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

  80. TuringTest
    • 2 years ago
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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?

  81. Inopeki
    • 2 years ago
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    division?

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

  83. Inopeki
    • 2 years ago
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    y with 3?

  84. TuringTest
    • 2 years ago
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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...?

  85. TuringTest
    • 2 years ago
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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)

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

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

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

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

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

  91. TuringTest
    • 2 years ago
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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...

  92. Inopeki
    • 2 years ago
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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

  93. TuringTest
    • 2 years ago
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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 ???

  94. Inopeki
    • 2 years ago
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    1.5?

  95. TuringTest
    • 2 years ago
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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?

  96. Inopeki
    • 2 years ago
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    3

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

  98. Inopeki
    • 2 years ago
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    1

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

  100. Inopeki
    • 2 years ago
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    0.3333333?

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

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

  103. TuringTest
    • 2 years ago
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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 :)

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

  105. TuringTest
    • 2 years ago
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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

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

  107. TuringTest
    • 2 years ago
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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

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

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

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

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

  112. TuringTest
    • 2 years ago
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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)

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

  114. TuringTest
    • 2 years ago
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    where?

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

  116. Inopeki
    • 2 years ago
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    Oh nevermind

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

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

  119. Inopeki
    • 2 years ago
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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.

  120. Inopeki
    • 2 years ago
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    flutter?

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

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

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

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

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

  126. TuringTest
    • 2 years ago
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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?

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

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

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

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

  131. TuringTest
    • 2 years ago
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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?

  132. Inopeki
    • 2 years ago
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    r

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

  134. TuringTest
    • 2 years ago
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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)=?

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

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

  137. TuringTest
    • 2 years ago
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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.

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

  139. Inopeki
    • 2 years ago
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    I see

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

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

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

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

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

  145. TuringTest
    • 2 years ago
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    right!

  146. Inopeki
    • 2 years ago
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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)

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

  148. TuringTest
    • 2 years ago
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    excellent!

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

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

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

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

  153. Inopeki
    • 2 years ago
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    As a point?

  154. TuringTest
    • 2 years ago
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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.

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

  156. TuringTest
    • 2 years ago
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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

  157. TuringTest
    • 2 years ago
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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

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

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

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

  161. TuringTest
    • 2 years ago
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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.

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

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

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

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

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

  167. Inopeki
    • 2 years ago
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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)

  168. TuringTest
    • 2 years ago
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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?

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

  170. TuringTest
    • 2 years ago
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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')

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

  172. TuringTest
    • 2 years ago
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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?

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

  174. TuringTest
    • 2 years ago
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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.

  175. Inopeki
    • 2 years ago
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    So now what?

  176. TuringTest
    • 2 years ago
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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?

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

  178. Inopeki
    • 2 years ago
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    NO! WAIT!

  179. TuringTest
    • 2 years ago
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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?

  180. Inopeki
    • 2 years ago
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    Oh...

  181. TuringTest
    • 2 years ago
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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)=?

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

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

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

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

  186. TuringTest
    • 2 years ago
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    yep :)

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

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

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

  190. TuringTest
    • 2 years ago
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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!

  191. Inopeki
    • 2 years ago
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    Thanks!

  192. TuringTest
    • 2 years ago
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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

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

  194. TuringTest
    • 2 years ago
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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!

  195. TuringTest
    • 2 years ago
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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

  196. Inopeki
    • 2 years ago
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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

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

  198. TuringTest
    • 2 years ago
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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!

  199. Inopeki
    • 2 years ago
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    Thanks :)

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

  201. Inopeki
    • 2 years ago
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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?

  202. Inopeki
    • 2 years ago
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    TT?

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