Solving Equations by Substitution Method

please help me out here, I know the answer but I don't know what the solution is D:

1) xy=15
x^2+2y^2=59 (the answer here is 3 and 5 but I dont know the solution)


2) a^2+b^2=41
a-b=1

1/
x=15/y
---> (15/y)^2 + 2y^2 = 59 , solve for y , then x

2/
a = b+1 ---> (b+1)^2+b^2=41 , solve for b , then a

xy = 15
x = 15/y

(15/y)^2 + 2y^2 = 59
(225/y^2) + 2y^2 = 59
225 + 2y^4 = 59y^2
2y^4 - 59y^2 + 225 = 0
(2y^2 - 9)(y^2 - 25) = 0

2y^2 = 9
y^2 = 9/2
y = ± √(9/2)
y = ± 3 / √2
y = ± 3√2 / 2

y^2 = 25
y = ± √25
y = ± 5


x(3√2 / 2) = 15
x = 30 / 3√2
x = 10 / √2
x = 10√2 / 2
x = 5√2

x = -5√2

x = 3
x = -3

(3, 5), (-3, -5), [5√2, (3√2 / 2)], [-5√2, (-3√2 / 2)]