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How to Solve: 1/4x + 2/9 - 14 = 100

  1. Jun 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Solve equation for x above.

    2. Relevant equations
    See title.

    3. The attempt at a solution

    Step 1: add 14 to both sides

    1/4x + 2/9 = 114

    Step 2: This is where my memory is a bit fuzzy.

    I know we can obtain like bases between 1/4x and 2/9, but that wouldn't seem super helpful, because they aren't like terms is that correct? I figured 1/4 and 2/9 would be like terms, but 1/4x and 2/9 aren't and cannot be combined.

    If that's the case, then it seems I'd have to deal with each fraction in both terms separately and one-by-one.

    I chose to "get rid" of the 2/9 fraction first by multiplying it by 9/1 (as well as ever other term on both sides by the same 9/1) and got:

    9/4x + 2 = 1026

    Step 3: Subtract 2 from both sides

    9/4x = 1024

    Step 4: Get rid of fraction by multiplying both sides by 4/1 and this yields:

    9x = 4096


    Step 5: Divide both sides by 9 to isolate the x and get:

    x = 455.11111111...

    Not 100% sure I did this correctly. And wondering also if there was a different or easier way?
     
  2. jcsd
  3. Jun 20, 2016 #2

    ProfuselyQuarky

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    You're main object is to get ##x## by itself. Regarding step #2, you can treat ##\frac{2}{9}## the same way you treated ##-14##. They are like terms, after all.
     
  4. Jun 20, 2016 #3
    Is your equation [itex]\frac{1}{4x}[/itex] or [itex]\frac{x}{4}[/itex] ? If it's the former then you have done it incorrectly
     
  5. Jun 20, 2016 #4

    ProfuselyQuarky

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    I believe that he was implying the former.
     
  6. Jun 20, 2016 #5
    Then his last step his wrong , if he multiplies both sides by 4/1 he would get [itex]\frac{9}{x} = 4096[/itex] . Not the other way.
     
  7. Jun 20, 2016 #6
    Hi, Mastermind

    1/4 is a coefficent of x. Apologies, b/c I don't know how to use the symbols stuff here yet. So, the x is NOT in the denominator of the first fraction.
     
  8. Jun 20, 2016 #7

    ProfuselyQuarky

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    In that case, I believe that you're correct.
     
  9. Jun 20, 2016 #8
    Right. You have done it correctly then.

    P.S : Here's a guide to Latex which implements the math symbols here: https://www.physicsforums.com/threads/introducing-latex-math-typesetting.8997/ This will help avoid future confusion.
     
  10. Jun 20, 2016 #9
    Oh, good to know I'm correct. Very rusty with certain topics that I'm trying to plug leaks in this summer.

    I am aware of LaTex, but haven't had time to really delve into learning it yet. I want to do so this summer though! I believe it will help me tremendously next year.
     
  11. Jun 20, 2016 #10
    Good luck then!
     
  12. Jun 20, 2016 #11
    Wait!

    Did you mean my last step was wrong even IF I meant 1/4 as the coefficient of x, instead of x being in the denominator?

    The two options you presented earlier weren't what I had meant. So just double checking one last time! Thanks again!
     
  13. Jun 20, 2016 #12
    No worries. Your last step is wrong only if x is in the denominator. Else it's correct.
     
  14. Jun 20, 2016 #13
    gotcha, thx
     
  15. Jun 20, 2016 #14

    SammyS

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    Then use parentheses to remove all doubt.

    What you wrote, 1/4x , literally means (1/4)x and is what you intended. However, if you include the parentheses, you remove all doubt regarding what you intended.
     
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