1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

3 unknowns really confusing

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Hello all,

    I'm trying to solve for 3 unknowns x,y,z. We are given these formulas: http://screencast.com/t/Y8QobUXVB3S3 [Broken]

    2. Relevant equations

    Please see link provided above.

    3. The attempt at a solution

    I first rearrage for root(x) then I rearrange equation 2 for root(y), I then sub root(x) into equation 2. From this point I'm a little confused on how to proceed?

    The problem I think is that in the formula for root(y) I still have root(z) unknown, root(z) has root(x) and also has a root (y), we know root(x) but it, itself contains root(y) and root(z) which is going in a weird loop that is making me really confused.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 21, 2013 #2
    Assuming you're treating ##\theta## and the various a's, b's, c's and d's as knowns, this system has nine unknowns, not three: ##x_1, x_2, x_3, y_1, y_2, y_3, z_1, z_2, z_3##.
     
  4. Jan 21, 2013 #3
    Hi LastOneStanding,

    I'm really sorry you are completly right, I wrote the equation wrong it should be: http://screencast.com/t/Y8QobUXVB3S3 [Broken]

    Although I still have the problem getting my head around that logic I mentioned before.

    Thanks!
     
    Last edited by a moderator: May 6, 2017
  5. Jan 21, 2013 #4
    Do you know to solve a linear system of equations using the elimination method? If not, see here for an explanation. I know this doesn't look linear because of the square roots, but it is linear if you just treat the square roots as the things you are trying to solve for instead of the things inside the square roots. Of course, once you've solved for each square root you can just square it to get the answer you really want.
     
  6. Jan 22, 2013 #5
    Hi LastOneStanding,

    Thanks very much that method does seem more efficient but I think we're supposed to do this is through substitution as thats the method our teacher has been using in class. Although I'm sure it will be a lot messier.

    So what would be the best way to approach this using the method of substituting in the values?

    Thanks!
     
  7. Jan 22, 2013 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If it is the square roots that bother you just replace them with, say, [itex]u= \sqrt{x}[/itex], [itex]v= \sqrt{y}[/itex] and [itex]w= \sqrt{z}[/itex]. Then you have three linear equations of the form
    [itex]A_1= B_1u+ C_1v+ D_1w[/itex]
    [itex]A_2= B_2u+ C_2v+ D_2w[/itex]
    [itex]A_3= B_3u+ C_3v+ D_3w[/itex]
    where I have also replaced the coefficients by single letters- you can put the coefficients back in after solving.

    Solve the first equation for u:
    [tex]u= \frac{A_1- C_1v- D_1w}{B_1}[/tex]

    and replace u in the other two equations by that:
    [itex]A_2= B_2\frac{A_1- C_1v- D_1w}{B_1}+ C_2v+ D_2w[/itex]
    [itex]A_3= B_3\frac{A_1- C_1v- D_1w}{B_1}+ C_3v+ D_3w[/itex]

    Solve either one of those for v and substitute into the other equation to get a single equation in w. Solve that equation for w, and substitute into the equation for v, the substitute both of those values into the original equation for u.

    Finally, of course, square u, v, and w to get x, y, and z.
     
  8. Jan 24, 2013 #7
    Thanks HallsOfIvy,

    Does this look correct to you?

    I mixed the terms up so now C if F and D is Z

    New equation link:
     
    Last edited: Jan 24, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook