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Solving by substitution for multiple variables

  1. Nov 27, 2012 #1
    1. The problem statement, all variables and given/known data
    I have four equations and have four variables. I need to solve for each of the variables. I am having difficulty figuring out how to do this.
    My equations are here. http://imgur.com/EOA8I

    2. Relevant equations

    [itex]\frac{(x-h)^2}{a^2}[/itex] + [itex]\frac{(y-k)^2}{b^2}[/itex] = 1

    3. The attempt at a solution
    here i plugged in b to equation a. http://imgur.com/egFhv
    Now from there if I plug that a into either h or k I am still stuck with three variables again. Is there an easier way to do this and how could I use wolfram alpha to help solve this?
     
  2. jcsd
  3. Nov 28, 2012 #2

    Simon Bridge

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    Use capitol letters for all the quantities you know so you can get a decent idea of the form of those equations. That means tidy up: expand out all those brackets and group terms.

    But I suspect you went forward too far - go back a couple of steps to the quadratics in h and k.
     
  4. Nov 28, 2012 #3

    Simon Bridge

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    Just to be sure...
    This is what I'm reading:
    $$2h = -7.24\pm\sqrt{7.24^2-4\left ( \left (1-\frac{1.0816-2.08k+k^2}{b^2}\right )a^2-13 \right )}$$
    $$2k = 3.64\pm\sqrt{3.64^2 - 4\left ( \left (1-\frac{4.4521-4.22h+h^2}{a^2}\right )b^2-3.3124 \right )}$$
    $$a^2=\frac{0.3136+1.12h+h^2}{1-\frac{24.5025-9.9k+k^2}{b^2}}$$
    $$b^2=\frac{4.84-4.4k+k^2}{1-\frac{14.6689+7.66h+h^2}{a^2}}$$
    ... are your four equations.
     
  5. Nov 28, 2012 #4

    Simon Bridge

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    Use the last two equations to get two equations for a and b in terms of h and k.
    These two equations go into the third one ... rearrange to get an equation for k in terms of h alone ... sub that into the top equation.

    That's why I figure you are better off with the forms of the top two before you applied the quadratic formula.
     
  6. Nov 28, 2012 #5
    yes, however I changed the coordinates for equation h.
    Here are my equations with the new, correct equation for h.

    http://imgur.com/hwScn

    NOTE: the equations are not multiplied out like I did earlier, but it looks like you read them right.

    I also tried working with wolfram alpha with this and got some very strange answers
     
  7. Nov 28, 2012 #6
    when plugging b into the equation for a, I end up with as shown here, http://tinyurl.com/c2onxp7

    then I simplified that down to

    a = ((4.95-k)(4.83-h)(-0.56-h))/(2.2-k)
     
  8. Nov 28, 2012 #7

    SammyS

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    Your equations for h & k look like they came from quadratic equations.

    If so, what are the original equations that you were given?
     
  9. Nov 28, 2012 #8
    equation for h was (13.1044+7.24h + h^2)/a^2 + (1.0816-2.08k + k^2)/b^2 = 1

    equation for k was (4.4521 - 4.22h + h^2)/a^2 + (3.3124 - 3.64k + k^2)/b^2 = 1

    although i changed the coordinates in equation for h with a 0 to make it simpler. And I also didn't multiply out the squares in the "relevant equations" mentioned in the first post. I just left the (x-h)^2 and (y-k)^2 as it is.

    this gave me new values that didn't require quadratics, (using (0, 4.74) for the x,y in equation h). I was able to simplfy to these equations without quadratic formulas;

    h = sqrt(a^2-a^2(4.74-k)^2/b^2)

    k = 1.82 +/- sqrt(b^2-b^2(2.11-h)^2/a^2)

    Am I on the right track?
     
  10. Nov 28, 2012 #9

    Simon Bridge

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    $$a = \frac{(4.95-k)(4.83-h)(-0.56-h)}{2.2-k}$$... good: that is a in terms of h and k.
    Do the same the other way: plug a into the equation for b and rearrange.
    $$h^2=a^2 - a^2\frac{( 4.47-k )^2}{b^2}$$
    $$k=1.82 \pm \sqrt{ b^2-b^2\frac{(2.11-h)^2}{b^2} }$$

    Impossible to tell - you are jumping all over the place.

    I still think you should leave the quadratic in place - you do not need to go as far as taking square roots here... leave that for the final steps in the simultaneous equations. All you need at the start is the relationships, you don't need to go all the way to making one of the variables the subject.

    From the above you are going to have to use care with the ##\pm## in the equation for k.
    The next step is to use the relations you have of ##a^2## and ##b^2## in terms of h and k to eliminate a and b from the above equations.

    Note: if you click the "quote" button under my post, it will show you how I got the equations to look so nice and easy to read.
    It is not hard to learn and it helps a great deal - go for it.
     
    Last edited: Nov 28, 2012
  11. Nov 28, 2012 #10
    Thanks for the help. I found out my algebra is weaker than I thought. I am confused with how I can reduce all these divisions. For say equation A, I cannot see how to get all the a's on one side. For example, when b is subbed into a,


    Thanks for the help. I found out my algebra is weaker than I thought. I am confused with how I can reduce all these divisions. For say equation A, I cannot see how to get all the a's on one side. For example, when b is subbed into a,

    I tried writing it in here but the /frac part confused the hell out of me.
    http://www.wolframalpha.com/input/?i=sqrt%28%28-.56-h%29^2%2F%281-%284.95-k%29^2%2F%282.2-k%29^2%2F%281-%28-3.83-h%29^2%2Fa^2%29%29

    What can I do with those divisions? Is a division of a division able to be brought up to the top?

    like for example;
    $$1-\frac{(-3.83-h)^2}{a^2)}$$

    is on the very bottom, can I bring it up to the numerator with the (4.95-k)^2? How else would I simplify to get the a^2 on the left side.
     
    Last edited: Nov 28, 2012
  12. Nov 28, 2012 #11

    Simon Bridge

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    Place the sum in the denominator over it's own common denominator... eg.$$x=\frac{y}{1-\frac{a+y}{x^2}}\\
    \Leftrightarrow x=\frac{y}{\frac{1}{x^2}(x^2-a-y)}\\
    \Leftrightarrow x=\frac{x^2y}{x^2-a-y}$$... you should be able to do the rest from there:$$\Rightarrow x(x^2-a-y)=x^2y\\
    \Rightarrow x^3-(a+y)x -yx^2=0\\
    \Rightarrow x^2-yx-(a+y)=0\\
    \Rightarrow y=\frac{x^2-a}{x+1}$$... ready to substitute into some relation of the form ##x=f(y)##

    Note: when you are solving a system involving conic sections - it is often easier to do it in polar coordinates.
     
    Last edited: Nov 28, 2012
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