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Math Help - Solving for y when it appears on both sides of equation

  1. #1
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    Solving for y when it appears on both sides of equation

    I've gotten stuck trying to solve the equation below for y. It's a tricky one for me as my algebra skills are basic at best.

    I have managed to get somewhere but the answer that comes out of the final equation is about 17% error from starting equation when I plug some numbers into it.

    Have I missed something or is there a better way?

    Start with 2 equations:

    1. y^2=r^2+x^2 and
    2. y=\frac{wx}{r}-d
    Need to substitute equation 2 into 1 and solve for y. Then resubstitute and solve for x. (Only r, w, and d are known).

    I got to this:

    3. y=\frac{w}{r}(\sqrt{y^2-r^2})-d

    But this leaves me with y^2
    on the RHS under the square root. From here is were I am unsure.

    I managed to get to this:

    4. y=\{\frac{1}{1-(\frac{r}{w})^2}[(\frac{dr}{w})^2+r^2]\}^0^.^5

    This got rid of y and x from RHS however, when I input some values to check if it works I get a different value for y between equations 3 and 4.

    Values I used are:

    r=3
    w=12
    d=11
    x=4
    y=5
    (had to use a number for y in eq. 3 to check it works)

    Same numbers in eq.4 y = 4.341?

    Have I gone about this the right way? what am I missing?

    Thanks.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Solving for y when it appears on both sides of equation

    Quote Originally Posted by Cartermgg View Post
    I've gotten stuck trying to solve the equation below for y. It's a tricky one for me as my algebra skills are basic at best.

    I have managed to get somewhere but the answer that comes out of the final equation is about 17% error from starting equation when I plug some numbers into it.

    Have I missed something or is there a better way?

    Start with 2 equations:

    1. y^2=r^2+x^2 and
    2. y=\frac{wx}{r}-d
    Need to substitute equation 2 into 1 and solve for y. Then resubstitute and solve for x. (Only r, w, and d are known).

    I got to this:

    3. y=\frac{w}{r}(\sqrt{y^2-r^2})-d

    But this leaves me with y^2
    on the RHS under the square root. From here is were I am unsure.

    I managed to get to this:

    4. y=\{\frac{1}{1-(\frac{r}{w})^2}[(\frac{dr}{w})^2+r^2]\}^0^.^5

    This got rid of y and x from RHS however, when I input some values to check if it works I get a different value for y between equations 3 and 4.

    Values I used are:

    r=3
    w=12
    d=11
    x=4
    y=5
    (had to use a number for y in eq. 3 to check it works)

    Same numbers in eq.4 y = 4.341?

    Have I gone about this the right way? what am I missing?

    Thanks.
    [*] y^2=r^2+x^2 and[*] y=\frac{wx}{r}-d

    (\frac{wx}{r}-d)^2=r^2+x^2

    Now, expand ans solve for x.
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  3. #3
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    Re: Solving for y when it appears on both sides of equation

    Thank you, I will have a go at it this way.
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  4. #4
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    Re: Solving for y when it appears on both sides of equation

    Quote Originally Posted by Also sprach Zarathustra View Post
    [*] y^2=r^2+x^2 and[*] y=\frac{wx}{r}-d

    (\frac{wx}{r}-d)^2=r^2+x^2

    Now, expand ans solve for x.

    I've tried long and hard to solve this equation but I can only get so far. I think I need a bit more guidance.

    I have managed to get to this:

    x^2(1-\frac{w^2}{r^2})+2\frac{wxd}{r}=d^2-r^2

    but I can't get any further. How do I get rid of the second x on the LHS?

    Thanks
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  5. #5
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    Re: Solving for y when it appears on both sides of equation

    Subtract d^2-r^2 from both sides:

    \left(1-\dfrac{w^2}{r^2}\right)x^2 + \left(\dfrac{2wd}{r}\right)x - (d^2-r^2) = 0

    Although it looks complicated this is a quadratic equation in x with
    1. =1 - \dfrac{w^2}{r^2}
    2. = \dfrac{2wd}{r}
    3. =d^2-r^2


    I would strongly recommend using the formula to solve this one.
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  6. #6
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    Re: Solving for y when it appears on both sides of equation

    Quote Originally Posted by Cartermgg View Post
    1. y^2=r^2+x^2 and
    2. y=\frac{wx}{r}-d
    Just divide eq1 by eq2 to get right away:
    y = r(r^2 + x^2) / (wx - dr)
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  7. #7
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    Re: Solving for y when it appears on both sides of equation

    Quote Originally Posted by Wilmer View Post
    Just divide eq1 by eq2 to get right away:
    y = r(r^2 + x^2) / (wx - dr)
    That didn't eliminate the \displaystyle x... The point of solving the equations simultaneously is to get \displaystyle x, y in terms of the other variables only.
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  8. #8
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    Re: Solving for y when it appears on both sides of equation

    I was answering his opening problem:

    "I've gotten stuck trying to solve the equation below for y"
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