How do you solve the system #x^2+y^2-3x=8# and #2x^2-y^2=10#?

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Answer 1 · 2016-08-24T05:53:46

(1) :In #RR^2#, the Solns. are #(x,y)=(3,+-2sqrt2)#.

(2) : In #CC^2#, the Solns. are #(x,y)=(3,+-2sqrt2),(-2, +-isqrt2)#.

Adding the given eqns., we get, #3x^2-3x=18, or, x^2-x-6=0#.

#:. ul(x^2-3x)+ul(2x-6)=0#.

#:. x(x-3)+2(x-3)=0#.

#:. (x-3)(x+2)=0#.

#:. x=3, or, x=-2#.

From the second eqn., since, #y^2=2x^2-10#,

# x=3 rArr y^2=8 rArr y=+-2sqrt2#, and,

# x=-2 rArr y^2=-2, "which is not possible in" RR#.

These roots satisfy the given system of eqns.

Hence, the Solns. in #RR^2#are #(x,y)=(3,+-2sqrt2)#.

Note :-

In #CC, y^2=-2 rArr y=+-isqrt2#.

Therefore, in #CC^2#, the Solns. are,

#(x,y)=(3,+-2sqrt2), (-2, +-isqrt2)#.

Enjoy Maths.!