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 $x-y=0$

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