<PRE><B>Question 30544</B>
Solve the system:
x^2+y^2=20 ----(1)
x^2-2y^2=8 ----(2)
(1) - (2)
(x^2-x^2)+[y^2- (-2y^2)] = 20-8
0 +(y^2+2y^2) = 12
3y^2 = 12
y^2 = 12/3= 4
y = +2 or -2
Putting y^2 = 4 in (1)
x^2+4 = 20
x^2 = 20-4
x^2 = 16
x= +4 or -4
Answer: (4,2),(4,-2), (-4,2) and (-4,-2)
which is your choice 
(D) (4,2) (4, -2) 
    (-4,2) (-4, -2)


</PRE>