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Equation in fraction form

  1. Mar 30, 2008 #1
    1. The problem statement, all variables and given/known data
    (3x-5)/4 + (2-3x)/2 = (x-2)/5


    2. Relevant equations



    3. The attempt at a solution

    I tried this and got x= 6.3 reacurring.... but i don't think it is right...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 30, 2008 #2
    [tex]\frac{3x-5}{4}+\frac{2-3x}{2}=\frac{x-2}{5}[/tex]

    Multiply everything by 20

    [tex]5(3x-5)+10(2-3x)=4(x-2)[/tex]

    Take it from here ...
     
  4. Mar 30, 2008 #3
    Show your work, how did you get that answer?
     
  5. Mar 30, 2008 #4
    I multiplied everything by 40, not 20, and then went 10(3x-5) + 20(2-3x) = 8(x-2)
     
  6. Mar 30, 2008 #5
    Ok, keep showing your work and we'll look over it. I'm not going to do it for you.

    You can check if your solution is correct by plugging it into your original equation ... left should equal the right.
     
  7. Mar 30, 2008 #6
    I know. then I expanded everything to make it :
    30x-50+40-60x=8x-16

    Then I combined like terms to make it -30x-10=8x-16

    Then I moved like terms to make it: -38x=-6

    Which equals 6.3 reacuring.... is that right?
     
  8. Mar 30, 2008 #7
    Work looks fine, now check your solution.
     
  9. Mar 30, 2008 #8
    Why are you dividing 38 by 6? You're trying to solve for the x.
     
  10. Mar 30, 2008 #9
    Don't you divide by the number in front of x by the number after the equals sign to get the answer for x?
     
  11. Mar 30, 2008 #10
    No. For example:

    [tex]6x = 12[/tex]

    [tex]\frac{6x}{6}=\frac{12}{6}[/tex]

    [tex]x=2[/tex]

    This is really basic algebra.
     
  12. Mar 30, 2008 #11
    But that is exactly what I did.... -38x/-38= -6/-38
     
  13. Mar 30, 2008 #12
    6/38 is not 6.333.
     
  14. Mar 30, 2008 #13
    oh... i see what i did wrong... thanks for pointing that out...
     
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