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3 equations, 3 unknowns, how do i solve ?

  1. Jan 27, 2006 #1
    3 equations, 3 unknowns, how do i solve!!!?

    hi everyone, i have 3 equations and 3 unknowns:

    x = (a/b) + c
    y = (a/(b+1)) + c
    z = (a/(b+7)) + c

    where x, y, z are given in the problem (real #'s), and we need to find the values of a, b, and c.

    can anyone tell me how to go about this problem? i tried rigourous algebra of solving and substituting but it became too difficult to isolate one of the variables a, b, or c.

    thank you in advance!
  2. jcsd
  3. Jan 27, 2006 #2


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    Homework Helper

    Multiply both sides by the denominator of the fraction. Then, define d=bc. You'll have three linear equations for a,b,d, and at the end you can get c from d and b.
  4. Jan 27, 2006 #3
    hmm i'm not sure i follow. so would i have xb = a + bc ? and then replace bc with d? how do i go from there then when i have xb = a + d?

  5. Jan 27, 2006 #4


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    Right, do that for all three equations, and then you'll have linear equations (since x,y,z are constants). These should be easy to solve. If you know linear algebra, that helps, but if not, it's just straightforward eliminating variables.
  6. Jan 27, 2006 #5
    okay so i have

    bx = a + d
    by + y = a + d + c
    bz + 7z = a + d + 7c

    where x,y,z are #'s and d=bc...
    if i want to put this in a matrix, is it possible? because i see the b's on the left side of the equation that are unknown...

    thank you for your patience.
  7. Jan 27, 2006 #6


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    Sorry, I kinda rushed through this before and made a mistake. You probably didn't need to introduce d. But it doesn't really matter, and it'll still work if you use the first equation to plug in bx for a+d in the other two equations. This will eliminate a in both of them, and that will leave you two linear equations for b and c. Just group all the b's and c's on one side.
    Last edited: Jan 27, 2006
  8. Jan 27, 2006 #7
    so now i have two linear equations:

    by - bx = c -y
    bz - bx = 7c - 7z

    which i'm not sure how set up from there as a linear eqn...should i factor a b out of the left side of the first equation then divide? please let me know!

  9. Jan 27, 2006 #8


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    Staff: Mentor

    You now have two equations in two unknowns (b and c). You've done that sort of thing before, right?

    For example, you could solve one equation for c and substitute it for c in the other one. Then you have a single equation for b only. Solve it. Then substitute your solution for b back into one of those two equations. Either one will work. Solve for c. Finally, take your solutions for b and c and substitute them back into one of your original equations. Any one of them will work. Solve for a.
  10. Jan 27, 2006 #9


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    Homework Helper

    You know, x, y, z. Rearrange a bit gives:
    [tex]\left\{ \begin{array}{l} (x - y)b + c = y \\ (x - z)b + 7c = 7z \end{array} \right.[/tex]
    Now that you have your matrix. You can go from here, right? :smile:
  11. Jan 27, 2006 #10
    thank you all for your help--it's appreciated!!! :!!)
  12. Jan 27, 2006 #11
    Be double careful that you don't multiply by zero and stumble across a solution that actually does not exist.
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