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How Do You Find X For This?

  1. Nov 18, 2009 #1
    How Do You Find X For This??

    Ok, how do you do problems like this and simplify?

    -7(1+-4m)=13(2m+-3)
    ??
    Also after you describe how to do it and stuff please give me a problem to work on and see if I'm right! Thanks

    please show and describe how you did it, and I'm in 8th grade so please talk so I can understand it hahah thanks again !!
     
  2. jcsd
  3. Nov 19, 2009 #2

    CompuChip

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    Re: How Do You Find X For This??

    Basically, it is just an equation of the form am + b = 0. Do you know how to solve these? For example, if I say:
    3m + 2 = 0
    or
    8m - 4 = 0
    then you say: m = ... ?

    Alternatively, you may have seen it in the form am = b, for example:
    3m = -2
    or
    8m = 4.

    The trick here is to rewrite a complicated equation like you have, into this form. The recipe is: first open up the brackets, for example: 3(2m + 4) = 3*2m + 3*4 = 6m + 12. In your example this is somewhat tricky, because you have all sorts of minus signs floating around which you should keep good track of (remember: positive times positive = negative times negative = positive, while negative times positive = positive times negative = negative).
    Once you did that, you get an equation like 3m + 4 = 8m - 9. You can bring this in the form you want by what I call the balance method: adding and subtracting on both sides of the equal sign. For this example (there are many many ways to do it)
    3m + 4 = 8m - 9 --- first add 9 on both sides
    3m + 4 + 9 = 8m - 9 + 9 --- the nines on the right hand side cancel
    3m + 13 = 8m --- now subtract 3m on both sides
    3m + 13 - 3m = 8m - 3m --- simplify: 3m on the left hand side cancels
    13 = (8-3)m --- so
    5m = 13

    You can solve this as m = 13/5. Or, if you want, you can again subtract 13 from both sides and write it as
    5m - 13 = 0.

    As you are see there are multiple steps involved. Since I don't really know where your problem with this exercise is, I will leave it to you now. Please give it a try and post your results here, so we can see where exactly you get stuck.
     
  4. Nov 19, 2009 #3
    Re: How Do You Find X For This??

    Well I can get to the last part but if like 43=60 what would the answer be then?
     
  5. Nov 19, 2009 #4

    Mark44

    Staff: Mentor

    Re: How Do You Find X For This??

    In general, if you end up with an impossible equation such as 43 = 60, and all of your steps leading up to this result are correct, then it means that there is no solution.

    For the particular problem you posted in this thread, though, it means that you did something wrong, because there is a solution to the equation you posted.
     
  6. Nov 20, 2009 #5
    Re: How Do You Find X For This??

    Ok. But what if there is one like

    #+n+#=n+#+n? Now how would you do that. That's the really hard stuff that I fail on. Because I take away the opposite amount of the number on the right side and add the the other just number. Then I take away from the n on the left and take away from the others. Then add and simplify? Right? Also, I forget what the problem looked like but do you ever take away from n and then would you take more from another n? I did that in class but I forget how. Thanks for helping me btw :D
     
  7. Nov 20, 2009 #6

    ideasrule

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    Re: How Do You Find X For This??

    I think you have the right idea, although I don't understand some of what you said.

    If you have something like #+n+#=n+#+n, add/subtract numbers or n's until you get something like 3n=5. Then you can easily solve for n.

    This applies to all equations. You can do ANY operation to both sides of the equation and the equation will still be valid. (After all, the two sides are equal.) To solve for a variable, you have to figure out how to get the equation into the form n=?
     
  8. Nov 20, 2009 #7

    CompuChip

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    Re: How Do You Find X For This??

    You can always "group together" similar quantities. For example,

    3 + 2n + 6 - 7n + 10n - 2 = 2n - 7n + 10n + 3 + 6 - 2
    ... = (2 - 7 + 10)n + (3 + 6 - 2)
    ... = 5n + 7 -- (1)

    So an equation like
    3 + 2n + 6 - 7n + 10n - 2 = 0 -- (2)
    would reduce to equation (1):
    5n + 7 = 0
    and the solution is then
    5n = -7 (subtract 7) -- (3)
    n = -7/5 (divide by 5)

    Similarly, if your equation was
    3 + 2n + 6 = 7n - 10n + 2
    you could reduce it to equation (2) by consecutively subtracting 7n, adding 10n, and subtracting 2 from both sides. Or you can first rewrite it as
    (2n) + (3 + 6) = (7n - 10n) + (2) -- group together
    2n + 9 = -3n + 2 -- now add 3n on both sides
    5n + 9 = 2
    Subtract 9 on both sides to get 5n = -7, or subtract 2 to get 5n + 7 = 0 -- getting you back to equation (1) or (3).

    And in reply to an earlier response: the type of equation which just contains one variable (n, m or x) and numbers in the way that you are posting, is called "linear" mathematics: being sums and differences of just numbers # and multiples #n of the variable). Now any linear equation always has one and precisely one solution, unless the variable cancels out such as in
    3n + 2 = 3n + 4 -- subtract 3n from both sides
    2 = 4 -- there is no n which satisfies this equation

    Otherwise, you can always bring it in the form
    #n = #
    by taking all n's to one side and all numbers to the other and dividing by the number that sits in front of the n.
     
    Last edited: Nov 20, 2009
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