Question

Solve the equations algebraivally, test the solution for validity, and graph the system to check your number of solutions.

x^2-2y^2=2
xy=2

Answer 1

so y = 2/x

x^2 - 2 (2/x)^2 = 2

x^2 - 8 /x^2 = 2

x^4 - 2 x^2 = 8
let z = x^2

z^2 - 2 z - 8 = 0

(z -4)(z+2)= 0

z = +4 or z = -2
so
x^2 = +4 or x^2 = -2
x = +2 then y = 1
x= -2 then y = -1
x = 2 i then y = 2/2i = -i
x = -2i then y = +i
put these all back in to check

Answer 2

What, no ideas of your own at all? This isn't even calculus; just algebra.

add them up to get

x^2+xy-2y^2 = 4
(x+2y)(x-y) = 4
just by inspection, (2,1) is a solution, and since both graphs are symmetric about (0,0), (-2,-1) is also a solution.

Algebraically, I guess you could just substitute y=2/x and get

x^2 - 2(4/x^2) = 2
x^4 - 8 = 2x^2
x^4-2x^2+1 = 9
(x^2-1)^2 = 9
x^2 = 1±3
x = ±2, ±√2 i