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Solve the simultaneous equations:

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$x+2y=\\var{x+2y}$

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$x^2+y^2=\\var{x^2+y^2}$

Rearranging the first equation gives $x=\\var{x+2y}-2y$

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Substituting this expression for $x$ into our second equation gives $(\\var{x+2y}-2y)^2+y^2=\\var{x^2+y^2}$

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Expanding brackets gives $\\simplify[]{{(x+2y)^2}-{4*(x+2y)}}$ $y+4y^2+y^2=\\var{x^2+y^2}$

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Rearranging gives $\\simplify[]{5y^2-{4*(x+2y)}y-{x^2+y^2-(x+2y)^2}}=0$

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Solving by factorising or using the quadratic formula gives $y=\\frac{\\var{4a} \\pm \\sqrt{\\var{16a^2+20(b-a^2)}}}{10}$

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Hence $y = \\var{y1}$ or $\\var{y2}$

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and the corresponding values of $x$ are $\\var{x1}$ and $\\var{x2}$ respectively (found by substituting each value of $y$ into $x+2y=\\var{x+2y}$)

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Lower value of $y=$[[0]]

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Corresponding value of $x=$[[1]]

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Greater value of $y=$[[2]]

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Corresponding value of $x=$[[3]]