Question 981319
x-y^2+6=0
2x^2+y^2-7=0, editing


x=y^2-6;x^2=y^4-12y^2+36
substitute
2y^4-24y^2+72+y^2-7=0

2y^4-23y^2+65=0
(2y^2-13)(y^2-5)=0
2y^2=13
y^2=(13/2)
y=  +/- sqrt  (26)/2; after rationalizing the denominator

y^2=5
y= +/- sqrt (5)

first equation becomes
x-5+6=0; x+1=0, x=-1

x=-1; y=+/- sqrt (5); +/- sqrt (26)/2

second equation becomes
2+5-7=0 ; checks