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#.