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Inopeki Group Title

Can someone give me a only letter equation to solve?

  • 2 years ago
  • 2 years ago

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  1. across Group Title
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    2x+3=11

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

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

    • 2 years ago
  4. Inopeki Group Title
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    Can i have a harder one now?

    • 2 years ago
  5. jimmyrep Group Title
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    good job

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

    • 2 years ago
  7. Inopeki Group Title
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    Nope but i take whats coming lol

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

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

    • 2 years ago
  10. across Group Title
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    That's right thus far!

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

    • 2 years ago
  12. Inopeki Group Title
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    ?

    • 2 years ago
  13. FoolForMath Group Title
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    That's correct.

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

    • 2 years ago
  15. Inopeki Group Title
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    Please put that in equation form

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

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

    • 2 years ago
  18. Inopeki Group Title
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    Hmm

    • 2 years ago
  19. Inopeki Group Title
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    Im supposed to "equalize the fractions" right?

    • 2 years ago
  20. Inopeki Group Title
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    Wait!

    • 2 years ago
  21. across Group Title
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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!\[\]

    • 2 years ago
  22. Inopeki Group Title
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    NONONONO

    • 2 years ago
  23. Inopeki Group Title
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    no clues :(

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

    • 2 years ago
  25. Inopeki Group Title
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    No that cant be it..

    • 2 years ago
  26. Inopeki Group Title
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    wait, just say yes or no, not why.

    • 2 years ago
  27. across Group Title
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    You rewrote the equation. :P

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

    • 2 years ago
  29. across Group Title
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    Yes

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

    • 2 years ago
  31. Inopeki Group Title
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    If so i have the answer

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

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

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

    • 2 years ago
  35. Inopeki Group Title
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    5-5 ---- 4-2?

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

    • 2 years ago
  37. Inopeki Group Title
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    :( i thought i knew this stuff...

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

    • 2 years ago
  39. Inopeki Group Title
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    Whats the logic in that?

    • 2 years ago
  40. across Group Title
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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).

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

    • 2 years ago
  42. across Group Title
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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:\[\frac{3+5}{2}=\frac{8}{2}=4.\]

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

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

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

    • 2 years ago
  46. Inopeki Group Title
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    yeah

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

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

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

    • 2 years ago
  50. across Group Title
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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?

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

    • 2 years ago
  52. Inopeki Group Title
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    You mean

    • 2 years ago
  53. Inopeki Group Title
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    make them higher so they match or something?

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

    • 2 years ago
  55. Inopeki Group Title
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    Oh you did mean like that

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

    • 2 years ago
  57. Inopeki Group Title
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    Aw man.. im really worthless at this..

    • 2 years ago
  58. Inopeki Group Title
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    At this rate ill never learn quantum physics..

    • 2 years ago
  59. across Group Title
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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.

    • 2 years ago
  60. Inopeki Group Title
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    Well how? i cant go upward.

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

    • 2 years ago
  62. Inopeki Group Title
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    Now i can have (15)/(4)

    • 2 years ago
  63. across Group Title
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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:\[\frac{35}{15}+\frac{12}{15}=\frac{47}{15}.\]

    • 2 years ago
  64. Inopeki Group Title
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    So i wasnt right?

    • 2 years ago
  65. across Group Title
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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?

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

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

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

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

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

    • 2 years ago
  71. across Group Title
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    Yes we can. :)

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

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

    • 2 years ago
  74. Inopeki Group Title
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    Is it right? :D

    • 2 years ago
  75. across Group Title
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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?

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

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

    • 2 years ago
  78. Inopeki Group Title
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    Please say its right..

    • 2 years ago
  79. across Group Title
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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:\[\frac{2}{3}x\implies3\cdot\frac{2}{3}x\implies\not{3}\cdot\frac{2}{\not{3}}x\implies2x.\]

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

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

    • 2 years ago
  82. Inopeki Group Title
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    5/5?

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

    • 2 years ago
  84. Inopeki Group Title
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    so essentially, i was right?

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

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

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

    • 2 years ago
  88. across Group Title
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    That's correct.

    • 2 years ago
  89. Inopeki Group Title
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    x=11/4

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

    • 2 years ago
  91. across Group Title
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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:\[-\frac{5}{4}\cdot\frac{3}{2}=?\]

    • 2 years ago
  92. Inopeki Group Title
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    -5/4*6/4 so 1/4?

    • 2 years ago
  93. across Group Title
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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!\[\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}.\]

    • 2 years ago
  94. Inopeki Group Title
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    -30/4?

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

    • 2 years ago
  96. Inopeki Group Title
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    Wait, do you know what this means?

    • 2 years ago
  97. across Group Title
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    What exactly?

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

    • 2 years ago
  99. across Group Title
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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!

    • 2 years ago
  100. across Group Title
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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.

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

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

    • 2 years ago
  103. LagrangeSon678 Group Title
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    and you will never become a physicist

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

    • 2 years ago
  105. Inopeki Group Title
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    Fumble you lagrangeson.

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

    • 2 years ago
  107. Inopeki Group Title
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    You suck. At least across tries to help me.

    • 2 years ago
  108. LagrangeSon678 Group Title
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    your saying i havent helped you?

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

    • 2 years ago
  110. LagrangeSon678 Group Title
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    prove it then you little squirt

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

    • 2 years ago
  112. LagrangeSon678 Group Title
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    we'll see ...

    • 2 years ago
  113. Inopeki Group Title
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    Damn right well see!

    • 2 years ago
  114. Inopeki Group Title
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    across, 15/6?

    • 2 years ago
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