How do you solve the system of $x^{2} + y^{2} = 8$ $x^2+y^2=8$ and $x y = 4$ $xy=4$ ?

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Answer 1 · 2015-03-24T16:43:42

An alternative solution method:

$x^{2} + y^{2} = 8$ $x^2+y^2=8$ and $x y = 4$ $xy=4$ is satisfied when:

$x^{2} + y^{2} = 2 x y$ $x^2+y^2=2xy$ . Which is true if and only if:

$x^{2} - 2 x y + y^{2} = 0$ $x^2-2xy+y^2=0$ .

Factoring, gives us:

${(x - y)}^{2} = 0$ $(x-y)^2=0$ , so $x = y$ $x=y$ .

Substituting in $x y = 4$ $xy=4$ gives us $x^{2} = 4$ $x^2=4$ , so $x = \pm 2$ $x=+-2$ .

Recalling that $y = x$ $y=x$ , the solutions are:

$x = y = 2$ $x=y=2$ and $x = y = - 2$ $x=y=-2$ .

How do you solve the system of x2+y2=8x^2+y^2=8 and xy=4xy=4?