I'm going out, so I shall do question 4 and 5 first. 1,2,3 use
substitution, similar to what I am going to do for 4 and 5
4)
2xy = -1
x = -1/2y
2x + 5y + 4 = 0
Sub x = -1/2y
-1/y + 5y + 4 = 0
5y^2 + 4y - 1 = 0
(5y - 1) (y + 1) = 0
y = 1/5 or y = -1
x = -5/2 or -1/2
5) 2y = x - 3
xy + 20 = 5x
2xy + 40 = 10x
Sub 2y = x - 3
x(x-3) + 40 = 10x
x^2 -13x + 40 = 0
(x-5) (x-8 ) = 0
x = 5 or 8
y = 1 or 2.5