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Math Help - [SOLVED] solving for x and making x the subject (yr 11 maths)

  1. #1
    Member smmmc's Avatar
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    solving for x and making x the subject (yr 11 maths)

    solving for x (i solved for y already, but subbing y into one of the equations seems to be too long of a process, is there another way?)
    a)x(b-c)+by-c=0 and y(c-a)-ax+c=0
    b)axy +b=(a+c)y and bxy+a=(b+c)y

    make x the subject
    x+a/x-c + x+c/x-a = 2 SOLVED


    thanks
    Last edited by smmmc; February 14th 2009 at 11:46 PM.
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  2. #2
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    Quote Originally Posted by smmmc View Post
    sol...

    make x the subject
    x+a/x-c + x+c/x-a = 2


    thanks
    I assume that you mean (x+a)/(x-c) + (x+c)/(x-a)=2

    If so:
    1. Multiply both sides of the equation by (x-c)(x-a). You'll get:

    (x+a)(x-a)+(x+c)(x-c)=2(x-c)(x-a)

    2. Expand the brackets:

    x^2-a^2+x^2-c^2=2(x^2-ax-cx+ab)

    3. Collect like terms. Move all terms containing x to the LHS, all terms with constants to the RHS:

    2ax+2cx=a^2+2ac+c^2~\implies~2(a+c)x=(a+c)^2

    4. Divide by the leading factor of x:

    x = \frac12\cdot (a+c)
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  3. #3
    Member smmmc's Avatar
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    Quote Originally Posted by earboth View Post

    3. Collect like terms. Move all terms containing x to the LHS, all terms with constants to the RHS:

    2ax+2cx=a^2+2ac+c^2~\implies~2(a+c)x=(a+c)^2
    thanks earboth. for the quoted above how did you get the (a+c)^2 ? from 2ac+c^2+a^2 ??
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  4. #4
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    Quote Originally Posted by smmmc View Post
    thanks earboth. for the quoted above how did you get the (a+c)^2 ? from 2ac+c^2+a^2 ??
    Correct. If you expand

    (a+c)^2=a^2+2ac+c^2

    I only used the reverse gear
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  5. #5
    Member smmmc's Avatar
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    ohhokay thanks. Anyone know of how to do the 2 remaining questions?
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  6. #6
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    Quote Originally Posted by smmmc View Post
    solving for x (i solved for y already, but subbing y into one of the equations seems to be too long of a process, is there another way?)
    a)x(b-c)+by-c=0 and y(c-a)-ax+c=0
    ...
    To solve such a system of simultaneous equations I would use the Cramer rule. If you have the system of linear equations:

    \begin{array}{l}a_1x+b_1y = c_1\\a_2x+b_2y = c_2\end{array}

    then the the solution is:

    x = \dfrac{c_1\cdot b_2 - c_2\cdot b_1}{a_1\cdot b_2 - a_2\cdot b_1} ..... and ........ y = \dfrac{a_1\cdot c_2 - a_2\cdot c_1}{a_1\cdot b_2 - a_2\cdot b_1}

    With you example you get:

    x = \dfrac{c^2-ac+bc}{ac+bc-c^2} ..... and ........ y = \dfrac{-bc+c^2+ac}{ac+bc-c^2} ........ which simplifies to:

    x = \dfrac{c-a+b}{a+b-c} ..... and ........ y = \dfrac{a-b+c}{a+b-c}
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  7. #7
    Member smmmc's Avatar
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    cramers rule uses matrixes right? i haven't learnt matrixes so this rule would be difficult for me..
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