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Math Help - simultaneous equations containing fractions

  1. #1
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    simultaneous equations containing fractions

    Hi;
    I have these equations 5/x - 2/y = 2 and 2/x + 3/y = 16.

    Unfortunatly I have no work to show because I don't know where to start with unknows on the bottom.

    Thanks
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    MHF Contributor Siron's Avatar
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    Re: simultaneous equations containing fractions

    Do you have to solve a system? Or are the two equations totally independent of each other? (I think the first one)
    In case you mean a sytem it can be useful to do a substitution, let \frac{1}{x}=a and let \frac{1}{y}=b (and x,y \neq 0), solve the system for a and b and afterwards do the back-substitution.
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    Re: simultaneous equations containing fractions

    Quote Originally Posted by anthonye View Post
    Hi;
    I have these equations 5/x - 2/y = 2 and 2/x + 3/y = 16.

    Unfortunatly I have no work to show because I don't know where to start with unknows on the bottom.

    Thanks
    x\neq 0 and y\neq 0.

    Substitute:

    \frac{1}{x}=t and \frac{1}{y}=u.
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    Re: simultaneous equations containing fractions

    Quote Originally Posted by anthonye View Post
    Hi;
    I have these equations 5/x - 2/y = 2 and 2/x + 3/y = 16.
    Here is a start.
    \left. \begin{gathered}  \frac{5}{x} - \frac{2}{y} = 2 \hfill \\  \frac{2}{x} + \frac{3}{y} = 16 \hfill \\ \end{gathered}  \right\} \equiv \left\{ \begin{gathered}  \frac{{15}}{x} - \frac{6}{y} = 6 \hfill \\  \frac{4}{x} + \frac{6}{y} = 32 \hfill \\ \end{gathered}  \right.
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    Re: simultaneous equations containing fractions

    Plato my result from here is 0=0 why.
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    Re: simultaneous equations containing fractions

    Quote Originally Posted by anthonye View Post
    Plato my result from here is 0=0 why.
    What do you mean with that?
    x=y\neq 0, because else you're diving by zero which is offcourse not possible (undefined).

    EDIT:
    If you use 'Plato's method' then make the sum of the two equations in the system therefore you'll get an equation in one variable which you can solve directly.
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    Re: simultaneous equations containing fractions

    Do I need to flip the fractions to solve them?
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    MHF Contributor Siron's Avatar
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    Re: simultaneous equations containing fractions

    Quote Originally Posted by anthonye View Post
    Do I need to flip the fractions to solve them?
    Make the sum of the two equations in the system, see my EDIT in my previous post.
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    Re: simultaneous equations containing fractions

    Quote Originally Posted by anthonye View Post
    Do I need to flip the fractions to solve them?
    Add these together \left\{ \begin{gathered}  \frac{{15}}{x} - \frac{6}{y} = 6 \hfill \\  \frac{4}{x} + \frac{6}{y} = 32 \hfill \\ \end{gathered}  \right.

    to get \frac{{19}}{x} = 38. Surely you can solve that.
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  10. #10
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    Re: simultaneous equations containing fractions

    So x = 0.5
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  11. #11
    MHF Contributor Siron's Avatar
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    Re: simultaneous equations containing fractions

    Quote Originally Posted by anthonye View Post
    So x = 0.5
    That's correct, substitute x=0,5 in one of the equations and solve the equation to y.
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  12. #12
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    Re: simultaneous equations containing fractions

    Ok thanks for the help just one more thing can they be solved by flipping them?
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  13. #13
    MHF Contributor Siron's Avatar
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    Re: simultaneous equations containing fractions

    Then you've to make a common denominator, if you want to flipp them then you've to write:
    \frac{5}{x}-\frac{2}{y}=2 \Leftrightarrow \frac{5y-2x}{xy}=2 \Leftrightarrow \frac{xy}{5y-2x}=\frac{1}{2}
    You see that this is not an easier way.
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    Re: simultaneous equations containing fractions

    Ok I'll look more later thanks for now.
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    MHF Contributor Siron's Avatar
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    Re: simultaneous equations containing fractions

    I will give you my solution:
    Let \frac{1}{x}=a and let \frac{1}{y}=b therefore you can write the equation as:
    \left\{ \begin{gathered} 5a - 2b = 2 \hfill \\  2a +3b  = 16 \hfill \\ \end{gathered}  \right.
    \Leftrightarrow \left\{ \begin{gathered} 15a - 6b = 6 \hfill \\  4a +6b  = 32 \hfill \\ \end{gathered}  \right.
    \Leftrightarrow 19a=38 \Leftrightarrow a=2

    If we substitute a=2 in one of the equations:
    10-2b=2 \Leftrightarrow 2b=8 \Leftrightarrow b=4

    Back-substitution:
    If a=2 \ \mbox{and} \ \frac{1}{x}=a \Rightarrow x=\frac{1}{2}
    y=\frac{1}{4}

    You can check your solutions by entering them into the sytem.
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