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Answer 1 · 2018-01-22T17:00:54

See below.

#{(x^2+y^2=25),(2x^2+6y^2=18):}#

Solving for #x^2, y^2# we have

#x^2=33# and #y^2 = -8# then no real solutions for this system of equations, the complex solutions being

#x = pmsqrt 32, y = pm i sqrt8#

Answer 2 · 2018-01-22T17:10:08

#(1):" The Soln. Set "sub RR" is "phi#.

#(2):" The Soln. Set "sub CC" is "x=+-sqrt33, y=+-2sqrt2i)#.

# x^2+y^2=25................<<1>>#.

# 2x^2+6y^2=18#.

Dividing by #2,# we get, #x^2+3y^2=9...........<<2>>#.

#:. <<2>> - <<1>> rArr 2y^2=-16#, which is not possible in #RR#.

Therefore, In #RR," the Solution (Soln.) Set is "phi#.

However, in #CC, 2y^2=-16 rArr y^2=-8 :. y=+-2sqrt2i#.

Then, fro #<<1>>, x^2+(-8)=25, i.e., x^2=33 :. x=+-sqrt33#.

To sum up,

#(1):" The Soln. Set "sub RR" is "phi#.

#(2):" The Soln. Set "sub CC" is "x=+-sqrt33, y=+-2sqrt2i)#.