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Math Help - two systems

  1. #1
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    two systems

    Solve each of the following two systems of equations


     (a) x + xy + y = 2+3\sqrt2,
     x^2+y^2 = 6



     (b) x^2 + y^2 +2xy/(x + y)=1,

     \sqrt{x+y} =x^2-y
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  2. #2
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    Hello,
    Quote Originally Posted by perash View Post
    Solve each of the following two systems of equations


     (a) x + xy + y = 2+3\sqrt2,
     x^2+y^2 = 6
    multiply the first one by 2 :
    2xy=4+6 \sqrt{2}-2(x+y)

    Add the two equations :
    \underbrace{x^2+y^2+2xy}_{(x+y)^2}=10+6 \sqrt{2}-2(x+y)

    (x+y)^2+2(x+y)=10+6 \sqrt{2}

    Complete the square :
    (x+y+1)^2=11+6 \sqrt{2}=(3+\sqrt{2})^2

    Hence x+y+1= \pm (3+\sqrt{2})

    x+y=\left\{\begin{array}{ll} -4-\sqrt{2} \quad (1) \\ 2+\sqrt{2} \quad (2) \end{array} \right.


    Let's do the first case : (1)
    Substitute in the equation x+xy+y=2+3 \sqrt{2} :
    xy=6+4 \sqrt{2}
    So you have the system :
    \left\{\begin{array}{ll} xy=6+4 \sqrt{2} \\ x+y=-4-\sqrt{2} \end{array} \right.


    For the second case : (2)
    Substitute x+y=2+\sqrt{2} in the very first equation :
    xy=2 \sqrt{2}
    So you have the system :
    \left\{\begin{array}{ll} xy=2 \sqrt{2} \\ x+y=2+\sqrt{2} \end{array} \right.

    Also, be careful of the extraneous solutions.


    Yah I know it's very long... I may have gone the wrong way ><
    The first system gives awful solutions, but the second system gives nice solutions : (2,sqrt(2)) and (sqrt(2),2)
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