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Inopeki

  • 3 years ago

Can someone give me a only letter equation to solve?

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  1. across
    • 3 years ago
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    2x+3=11

  2. Inopeki
    • 3 years ago
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    take 3 from both sides getting 2x=8 divide both sides by 2 gettin x=4

  3. Inopeki
    • 3 years ago
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    Btw, grats on Master rank

  4. Inopeki
    • 3 years ago
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    Can i have a harder one now?

  5. jimmyrep
    • 3 years ago
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    good job

  6. FoolForMath
    • 3 years ago
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    that's called only a letter equation? :D

  7. Inopeki
    • 3 years ago
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    Nope but i take whats coming lol

  8. across
    • 3 years ago
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    Let's do a slightly trickier one: (1/3)x+2=3

  9. Inopeki
    • 3 years ago
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    Ill take (1/3)x+2=3 1x+6=9, right sofar?

  10. across
    • 3 years ago
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    That's right thus far!

  11. Inopeki
    • 3 years ago
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    x+6=9 x=3

  12. Inopeki
    • 3 years ago
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    ?

  13. FoolForMath
    • 3 years ago
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    That's correct.

  14. across
    • 3 years ago
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    Very good! Let's do this one now: (2/3)x+(5/2)=(5/4) (I love fractions.)

  15. Inopeki
    • 3 years ago
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    Please put that in equation form

  16. across
    • 3 years ago
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    \[\frac{2}{3}x+\frac{5}{2}=\frac{5}{4}\]

  17. across
    • 3 years ago
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    If you can do that without problems, then you would have mastered the solving of single-variable, linear equations. :P

  18. Inopeki
    • 3 years ago
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    Hmm

  19. Inopeki
    • 3 years ago
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    Im supposed to "equalize the fractions" right?

  20. Inopeki
    • 3 years ago
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    Wait!

  21. across
    • 3 years ago
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    We're trying to solve for \(x\), so try and use the properties of addition, subtraction, multiplication and division to achieve that!\[\]

  22. Inopeki
    • 3 years ago
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    NONONONO

  23. Inopeki
    • 3 years ago
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    no clues :(

  24. Inopeki
    • 3 years ago
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    (2/3)x+(5/2)=(5/4)----->(4/6)x+(10/4)=(10/8)?

  25. Inopeki
    • 3 years ago
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    No that cant be it..

  26. Inopeki
    • 3 years ago
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    wait, just say yes or no, not why.

  27. across
    • 3 years ago
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    You rewrote the equation. :P

  28. Inopeki
    • 3 years ago
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    (2/3)x+(5/2)=(5/4) (2/3)x=(5/4)-(5/2) Right so far?

  29. across
    • 3 years ago
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    Yes

  30. Inopeki
    • 3 years ago
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    Now multiply both sides by 3 to get (3/3)x=(5/4)-(5/2)*3?

  31. Inopeki
    • 3 years ago
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    If so i have the answer

  32. across
    • 3 years ago
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    You have the right idea, but you're overlooking a few steps. :)

  33. Inopeki
    • 3 years ago
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    (3/3)x=(5/4)*3-(5/2)*3??

  34. across
    • 3 years ago
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    Let's first focus on trying to simplify this:\[\frac{5}{4}-\frac{5}{2}.\]Can you do that?

  35. Inopeki
    • 3 years ago
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    5-5 ---- 4-2?

  36. across
    • 3 years ago
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    There's where we're getting stuck. :P You need to learn how to simplify expressions of the form\[\frac{a}{b}\pm\frac{c}{d}.\]

  37. Inopeki
    • 3 years ago
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    :( i thought i knew this stuff...

  38. across
    • 3 years ago
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    It's not too bad, really, there's a simple rule which states that\[\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}.\]

  39. Inopeki
    • 3 years ago
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    Whats the logic in that?

  40. across
    • 3 years ago
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    To put it simply, you cannot add nor subtract fractions with different bases. In other words,\[\frac{5}{4}-\frac{5}{2}\]cannot be simplified as it stands since they have different bases (4 and 2).

  41. Inopeki
    • 3 years ago
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    So i need to find the greates common (something i cant remember)?

  42. across
    • 3 years ago
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    However, if you have something like\[\frac{3}{2}+\frac{5}{2},\]they have similar bases and thus you can add them accordingly:\[\frac{3+5}{2}=\frac{8}{2}=4.\]

  43. across
    • 3 years ago
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    You can do that OR you could use the fact that\[\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}.\]

  44. Inopeki
    • 3 years ago
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    How about if they have similar (the ones we have over the base)(whatever they call them)

  45. across
    • 3 years ago
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    The numerator? Do you mean something like\[\frac{3}{5}+\frac{3}{6}?\]

  46. Inopeki
    • 3 years ago
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    yeah

  47. Inopeki
    • 3 years ago
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    Oh wait, never mind i think i see why that wont work

  48. across
    • 3 years ago
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    It doesn't matter: they have different bases and you can't add them.

  49. Inopeki
    • 3 years ago
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    (2/3)x=(5/4)-(5/2) 5*2-4*5/2*5?

  50. across
    • 3 years ago
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    I saw something I liked: you know how to manipulate fractions. Could you manipulate\[\frac{5}{4}-\frac{5}{2}\]so that they have similar bases?

  51. Inopeki
    • 3 years ago
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    Im not sure i know what you mean by manipulate

  52. Inopeki
    • 3 years ago
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    You mean

  53. Inopeki
    • 3 years ago
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    make them higher so they match or something?

  54. across
    • 3 years ago
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    With "manipulate" I mean to say "play with" or "rearrange," like this:\[\frac{5}{2}=\frac{10}{4}=\frac{15}{6}=etc.\]

  55. Inopeki
    • 3 years ago
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    Oh you did mean like that

  56. Inopeki
    • 3 years ago
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    (5/2) (5/4)-------->2.5/2?

  57. Inopeki
    • 3 years ago
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    Aw man.. im really worthless at this..

  58. Inopeki
    • 3 years ago
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    At this rate ill never learn quantum physics..

  59. across
    • 3 years ago
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    That's good! But try and make the bases of the two fractions match and try to have only whole numbers in both the numerator and denominator.

  60. Inopeki
    • 3 years ago
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    Well how? i cant go upward.

  61. Inopeki
    • 3 years ago
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    (5/2)---->(10/4)!!!! (5/4)

  62. Inopeki
    • 3 years ago
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    Now i can have (15)/(4)

  63. across
    • 3 years ago
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    I'll give you an example. I'm given\[\frac{7}{3}+\frac{4}{5}.\]I want their denominators match, so I multiply \(7/3\) by \(5/5\) to get \(35/15\), and I multiply \(4/5\) by \(3/3\) to get \(12/15\). They now have similar denominators and I can add them:\[\frac{35}{15}+\frac{12}{15}=\frac{47}{15}.\]

  64. Inopeki
    • 3 years ago
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    So i wasnt right?

  65. across
    • 3 years ago
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    Very good! You converted\[\frac{5}{4}-\frac{5}{2}\]into\[\frac{5}{4}-\frac{10}{4}.\]What's the next step?

  66. Inopeki
    • 3 years ago
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    (2/3)x=(5/4)-(10/4) (2/3)x=(10-5)/4?

  67. across
    • 3 years ago
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    That's very close! But you switched the signs:\[\frac{5}{4}-\frac{10}{4}=\frac{5-10}{4}.\]

  68. Inopeki
    • 3 years ago
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    Oh right! so its (2/3)x=(-5)/4

  69. across
    • 3 years ago
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    That's correct! Now, let's solve for x. :)

  70. Inopeki
    • 3 years ago
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    Can we multiply both sides by 3 now?

  71. across
    • 3 years ago
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    Yes we can. :)

  72. Inopeki
    • 3 years ago
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    (3/3)x=(-5)/4*3

  73. Inopeki
    • 3 years ago
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    x=(-5)/4*3

  74. Inopeki
    • 3 years ago
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    Is it right? :D

  75. across
    • 3 years ago
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    Well, let's look at the LHS only: when you multiplied\[\frac{2}{3}x\]by \(3\), you somehow got\[\frac{3}{3}x.\]How did this happen?

  76. Inopeki
    • 3 years ago
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    Wait, should i divide both sides by 2/3?

  77. Inopeki
    • 3 years ago
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    x=(-5)/4*(2/3)

  78. Inopeki
    • 3 years ago
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    Please say its right..

  79. across
    • 3 years ago
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    You could do that, yes, but let's do it one step at a time first so that you understand the process. When you have\[\frac{2}{3}x,\]to get rid of that \(3\) in the denominator, we DO multiply the fraction by \(3\) so that this happens:\[\frac{2}{3}x\implies3\cdot\frac{2}{3}x\implies\not{3}\cdot\frac{2}{\not{3}}x\implies2x.\]

  80. Inopeki
    • 3 years ago
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    Oh right, i knew it had something to do with multiplying with the denominator!

  81. across
    • 3 years ago
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    Now, let's do a cool little trick: what happens if I multiply\[\frac{2}{3}x\]by\[\frac{3}{2}?\]:)

  82. Inopeki
    • 3 years ago
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    5/5?

  83. across
    • 3 years ago
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    Close!\[\frac{3}{2}\cdot\frac{2}{3}x=\frac{6}{6}x=x.\]

  84. Inopeki
    • 3 years ago
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    so essentially, i was right?

  85. across
    • 3 years ago
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    Now, can you apply this to solve\[\frac{2}{3}x=-\frac{5}{4}?\]

  86. across
    • 3 years ago
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    Yes, you were right, but I don't think you understood the whole process, which now you do. :)

  87. Inopeki
    • 3 years ago
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    Both sides multiplied by 3/2?

  88. across
    • 3 years ago
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    That's correct.

  89. Inopeki
    • 3 years ago
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    x=11/4

  90. Inopeki
    • 3 years ago
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    If thats not right, im gonna go give up on physics and go cry in a corner..

  91. across
    • 3 years ago
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    I think you got it already, but you're trying to speed your way through it. Take your time! :) Let's check again:\[-\frac{5}{4}\cdot\frac{3}{2}=?\]

  92. Inopeki
    • 3 years ago
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    -5/4*6/4 so 1/4?

  93. across
    • 3 years ago
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    I see what you're trying to do, but remember that multiplication and division of fractions is totally different from addition and subtraction. :) They're easier, in fact!\[\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}.\]

  94. Inopeki
    • 3 years ago
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    -30/4?

  95. across
    • 3 years ago
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    Look at our final expression:\[-\frac{5}{4}\cdot\frac{3}{2}=-\frac{5\cdot3}{4\cdot2}=?\]

  96. Inopeki
    • 3 years ago
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    Wait, do you know what this means?

  97. across
    • 3 years ago
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    What exactly?

  98. Inopeki
    • 3 years ago
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    That im a complete and utter failure and that i will never become a physicist.

  99. across
    • 3 years ago
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    Hey, it's your first time doing this. Don't beat yourself too hard. We're taking it one step at a time, and that's the best way there is to go!

  100. across
    • 3 years ago
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    Also, when I had your age, I knew much, much less than you do, and now look at me; I'm a math grad student! Your enthusiasm stretches beyond any of that I've seen in most students your age. Please keep at it; you will go really far. Take advantage of the fact that there are people out there that will gladly help you, and you can for sure include me in that group of people.

  101. wasiqss
    • 3 years ago
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    susanne its really sweet of u to help sumone like this with this devotion

  102. wasiqss
    • 3 years ago
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    n yehh that kid is damn focused on the job for sure!

  103. LagrangeSon678
    • 3 years ago
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    and you will never become a physicist

  104. Inopeki
    • 3 years ago
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    Ok, im done sulking. Wanna do some math?

  105. Inopeki
    • 3 years ago
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    Fumble you lagrangeson.

  106. LagrangeSon678
    • 3 years ago
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    those word came out of your mouth, fumble yourself

  107. Inopeki
    • 3 years ago
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    You suck. At least across tries to help me.

  108. LagrangeSon678
    • 3 years ago
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    your saying i havent helped you?

  109. Inopeki
    • 3 years ago
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    You just said i wasnt gonna become a physicist! Mark my words, I WILL!

  110. LagrangeSon678
    • 3 years ago
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    prove it then you little squirt

  111. Inopeki
    • 3 years ago
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    Screw you, ill get 2 phds

  112. LagrangeSon678
    • 3 years ago
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    we'll see ...

  113. Inopeki
    • 3 years ago
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    Damn right well see!

  114. Inopeki
    • 3 years ago
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    across, 15/6?

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