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

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Answer 1 · 2015-03-24T16:43:42

An alternative solution method:

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

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

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

Factoring, gives us:

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

Substituting in $xy=4$ gives us $x^2=4$ , so $x=+-2$ .

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

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

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