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Algebra II Word Problem Involving Fractional Equations

  1. May 1, 2012 #1
    Hi everyone! This time I have a lesson on word problems, and that is probably my absolute weakest point in math. (In fact, as I write, I'm puzzling over several different problems, but here is the one I'll post for now.)


    1. The problem statement, all variables and given/known data
    Members of the Computer Club were assessed equal amounts to raise $1200 to buy some software. When 8 new members joined, the per-member assessment was reduced by $7.50. What was the new size of the club?


    2. Relevant equations
    --


    3. The attempt at a solution
    I understand what it is asking for, and the information provided. My problem with word problems is translating the facts into a formula.

    I assigned d = dollars and m = the number of members

    The original amount of members and dollars would be:

    [tex]\frac{1200}{m} = d [/tex]

    Then, with the new members:
    [tex]\frac{1200}{m + 8} = d - 7.5 [/tex]

    This is my attempt. I'm not confident that I set the formula up correctly or how to solve it if it is correct. I could use some hints right about now. :redface: Please and thank you!
     
  2. jcsd
  3. May 1, 2012 #2
    The equations are correct. This is just a system of equations. You already have one variable expressed in therms of another. Now you only have to substitute that expression in the second equation and solve it. It will be a quadratic equation, when you clean it up.
     
  4. May 1, 2012 #3
    Remember that one of the solutions to the quadratic won't make sense and you should throw away that term.
     
  5. May 1, 2012 #4

    HallsofIvy

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    Excellent! And that is the same as 1200= md

    Good. So 1200= (m+8)(d- 7.5)= md+ 8d- 7.5m- 60= 1200+ 8d- 7.5m- 60
    which reduces to 8d- 7.5m= 60. You can solve that for d (or m) and put that into 1200= md.

     
  6. May 2, 2012 #5
    Ah ha ha! My math instincts aren't quite there yet. I thought it might be a system of equations, and even got as far as [itex]8d- 7.5m= 60 [/itex] but I second guessed it and tossed that idea out. :frown:Going with it then, I think I'm doing the math wrong.


    So, for starters, if I'm expressing 'd' in terms of 'm' is this correct: [itex]d =\frac {7.5m + 60}{8}[/itex]

    Or, expressing 'm' in terms of 'd' would be: [itex]d =\frac {-8d + 60}{7.5}[/itex]



    Are both of those correct? And you plug them back in for [itex]1200 = md[/itex] When I do that, I get a huge quadratic equation that is confusing to solve, such as [itex] 8d^2 -60d + 9000 [/itex]

    I also forgot to include the answer. Sorry, it's proof I was frazzled working on all those word problems! (I'll go back and edit it.) The answer is 40 members [of the new club].
     
  7. May 2, 2012 #6

    HallsofIvy

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    First, I can see no point in reducing to an equation for "d" because you don't need to know d to answer the question- it asks for the number of people in the club, not how much each has to pay.
    The second equation has a typo- you mean
    [tex]m= \frac{-8d+ 60}{7.5}[/tex]
    Yes, both are correct. Now, combine the first, where you solved for d, with 1200= md so you have
    [tex]1200= m\frac{7.5m+ 60}{8}[/tex]

    Multiplying both sides by 8, [itex]9600= 7.5m^2+ 60m[/itex] or [itex]7.5m^2+ 60m- 9600= 0[/itex]. If you don't like fractions, you can multiply both sides by 2 to get [itex]15m^2+ 120m- 19200= 0[/itex]. Dividing through by 5 gives [itex]3m^2+ 24m-3840= 0[/itex] and then dividing by 3, [itex]m^2+ 8m- 1280= 0[/itex].

    Solve that using the quadratic formula. Don't forget that the problem asks for the number of people in the new club- m+ 8.
     
  8. May 3, 2012 #7
    @HallsofIvy: Well, technically, it would be adding an extra step, but I could solve for 'd' first and then use that to find 'm', right?


    So factoring out, [itex] 7.5m^2 +60m−9600=0[/itex], I got [itex](1.5m - 48)(5m + 200)[/itex].

    One of which, as scurty said, isn't possible since you can't have negative people. And, so that leaves [itex]\frac {48}{1.5}[/itex], which equals 32.

    And so, if m = 32, then m + 8 = 40. So, there are 40 people in the new club. And that is the correct answer the book provides.

    So, was my method correct?
     
  9. May 3, 2012 #8

    HallsofIvy

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    Yes, that is correct. And I didn't say it was wrong to write the equation of d and then, after finding d, solve another equation for m. I just said that I did not see any point in doing that rather than just set up the equation in terms of m in the first place.
     
  10. May 3, 2012 #9
    @HallsofIvy: I understand. I was just double-checking.

    Thank you hamsterman, scurty, and HallsofIvy! I appreciate all of your help! :)
     
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