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Math Help - simultaneous equation

  1. #1
    Senior Member
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    simultaneous equation

    \frac{3x+y+1}{8}=\frac{x-y}{5}=\frac{x^2-y^2}{5}



    \frac{3x+y+1}{8}=\frac{x-y}{5}

    7x+13y=-5 --- 1
     <br /> <br />
\frac{x-y}{5}=\frac{x^2-y^2}{5}<br />

    x-y=(x+y)(x-y)

    x+y=1 (I am not sure if i can do this) ----2

    So solving 1 and 2 , i gt x=3 and y=-2 .

    It looks to me that there is another pair of values x and y but just couldn't find them . I am confusing my self at times . How can i find the other pair ?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by thereddevils View Post
    \frac{3x+y+1}{8}=\frac{x-y}{5}=\frac{x^2-y^2}{5}



    \frac{3x+y+1}{8}=\frac{x-y}{5}

    7x+13y=-5 --- 1
     <br /> <br />
\frac{x-y}{5}=\frac{x^2-y^2}{5}<br />

    x-y=(x+y)(x-y)

    x+y=1 (I am not sure if i can do this) ----2

    So solving 1 and 2 , i gt x=3 and y=-2 .

    It looks to me that there is another pair of values x and y but just couldn't find them . I am confusing my self at times . How can i find the other pair ?
    Hi

    Better factor : x-y=(x+y)(x-y)

    (x-y)(x+y-1)=0

    2 possibilities :
    or x+y-1 = 0 (you have solved this case)
    or x-y = 0 (you have forgotten this one)
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