How do you solve the system #x^2+y^2=18# and #x-y=0#?

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Answer 1 · 2016-07-26T13:42:30

#x=3,y=3# or #x=-3,y=-3#

Multiplying by #x+y# both sides of

#x-y=0#

results the following system

# { (x^2+y^2=18), (x^2-y^2 = 0) :}#

So #2x^2=18->x=pm 3# and #y = pm 3#

The feasible solutions are

#x=3,y=3# and #x=-3,y=-3#. Their feasibility was checked against the equation #x-y=0#