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Thread: Simplifying a simple algebra equation.

  1. #1
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    Simplifying a simple algebra equation.

    Hi guys,

    $\displaystyle A-(A-B)/(6B-5)*(6B-5)=B$

    Totally suck at math (haven't done it since school and barely passed) but I need to use some simple algebra equations in some stuff I'm doing at the moment, I would appreciate any help.

    My question is in the equation above I will know what number A is but I still don't know how to use it to find B.
    Example when A is 16 B is 2 but I don't know how to figure it out, could you simplify this equation so that I can find B easily if I know A?

    Thanks guys
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  2. #2
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    Quote Originally Posted by Not2l8 View Post
    Hi guys,

    $\displaystyle A-(A-B)/(6B-5)*(6B-5)=B$

    Totally suck at math (haven't done it since school and barely passed) but I need to use some simple algebra equations in some stuff I'm doing at the moment, I would appreciate any help.

    My question is in the equation above I will know what number A is but I still don't know how to use it to find B.
    Example when A is 16 B is 2 but I don't know how to figure it out, could you simplify this equation so that I can find B easily if I know A?

    Thanks guys
    Hi not2l8,

    Is this what you have written:

    $\displaystyle \frac{A-(A-B)}{6B-5} \cdot (6B-5)=B$

    or is it this:

    $\displaystyle A-\frac{A-B}{6B-5} \cdot (6B-5)=B$

    or something else?

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  3. #3
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    Quote Originally Posted by masters View Post


    $\displaystyle A-\frac{A-B}{6B-5} \cdot (6B-5)=B$

    or something else?
    Hi Masters, that is the one I was trying to write.
    I can see now though that they just cancel each other out.
    Mmmm... Doh, I guess that's not the equation I'm looking for.

    Thanks for looking at it though
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  4. #4
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    Quote Originally Posted by Not2l8 View Post
    Hi guys,

    $\displaystyle A-(A-B)/(6B-5)*(6B-5)=B$
    This would be

    $\displaystyle A-\left(\frac{A-B}{6B-5}\right)(6B-5)=B$

    but I am assuming that you intended the second $\displaystyle (6B-5)$ to be included in the denominator.

    $\displaystyle A-\frac{A-B}{(6B-5)(6B-5)}=B$

    $\displaystyle \Rightarrow A-\frac{A-B}{36B^2-60B+25}=B$

    $\displaystyle \Rightarrow\frac{A\left(36B^2-60B+25\right)-(A-B)}{36B^2-60B+25}=B$

    $\displaystyle \Rightarrow\frac{36AB^2-60AB+25A-A+B}{36B^2-60B+25}=B$

    $\displaystyle \Rightarrow36AB^2-60AB+24A+B=36B^3-60B^2+25B$

    $\displaystyle \Rightarrow36B^3-36AB^2-60B^2+60AB-24A+24B=0$

    $\displaystyle \Rightarrow36B^3-(36A+60)B^2+(60A+24)B-24A=0$

    Since this is a cubic in $\displaystyle B,$ there are at most 3 real solutions. Using the rational roots theorem, we may try $\displaystyle B=1,$ and using some synthetic division we do indeed find this to be a solution. The cubic then factors as

    $\displaystyle (B-1)\left[36B^2-(36A+24)B+24A\right]=0$

    Now we can simply use the quadratic formula. In the end, after factoring, we get

    $\displaystyle (B-1)(B-A)(3B-2)=0$

    Therefore, the solutions to the original equation are

    $\displaystyle \left\{\begin{array}{rcl}
    B&=&A\\
    B&=&1\\
    B&=&2/3\end{array}\right.$

    In case it isn't clear, $\displaystyle B=1$ and $\displaystyle \mbox{$B=\frac23$}$ are always solutions, regardless of the value of $\displaystyle A.$ The third solution is $\displaystyle B=A.$

    Edit: Okay, it seems that the problem was as you had written it. In that case, all values of $\displaystyle A$ and $\displaystyle B$ satisfy the equation.
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  5. #5
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    Thanks very much Reckoner, I must look like such a dunce asking such simple stuff

    But I 'think' I finally figured out what I'm trying to ask, I think it is this equation below -

    $\displaystyle
    (A-B)/(6B-5)=B$

    Only one or two values of B make it work I think.
    Like if A is 16 only B=2 makes the equation make sense.
    ((6B-5) is meant to the denominator in the equation above)

    So how do I simplify that so it is easy to find B if I know A
    Last edited by Not2l8; Apr 29th 2009 at 06:13 AM.
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  6. #6
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    Quote Originally Posted by Not2l8 View Post
    $\displaystyle
    (A-B)/(6B-5)=B$

    Only one or two values of B make it work I think.
    Like if A is 16 only B=2 makes the equation make sense.
    No, $\displaystyle A=16,\;B=-\frac43$ is another solution. You will never have more than two, though.

    So how do I simplify that so it is easy to find B if I know A
    $\displaystyle \frac{A-B}{6B-5}=B$

    $\displaystyle \Rightarrow A-B=B(6B-5)$

    $\displaystyle \Rightarrow A-B=6B^2-5B$

    $\displaystyle \Rightarrow 6B^2-4B-A=0$

    From the quadratic formula,

    $\displaystyle B=\frac{4\pm\sqrt{16+24A}}{12}$

    $\displaystyle =\frac{4\pm2\sqrt{4+6A}}{12}$

    $\displaystyle =\frac13\pm\frac{\sqrt{4+6A}}6$
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