solve each system by the substitution method. 1) 2x^2-y^2=6 2) y=x^2-3

 


 



 



so y=0 or y=2 



y=0: 

 


 

or 



 


y=2: 


or
 

solutions 

(-sqrt(3),0)
(sqrt(3),0)
(-sqrt(5),2)
(sqrt(5),2) 



:)