Substitutions: α = a+b; β = a-b; λ = a/b
Restrictions: α ≠ 0; β ≠ 0; x ≠ y

x   y       x+y
― + ― = 2 ٨ ――― = λ
α   β       x-y

Lets work each equation sep. in terms of x
-----------------------------
x   y
― + ― = 2
α   β

x       y
― = 2 - ―
α       β

                 α
Eq.1) x = 2α - y.―
                 β

-----------------------------
x+y
――― = λ
x-y

x+y
――― - λ = 0
x-y

x+y     x-y
――― - λ.――― = 0
x-y     x-y

x+y-(x-y).λ
――――――――――― = 0
    x-y

x+y-(x-y).λ = 0
x+y-x.λ+y.λ = 0
x(1-λ)+y(1+λ) = 0
x(1-λ) = -y(1+λ)
            y(1+λ)
Eq.2) x = - ――――――
             1-λ
-----------------------------
Combine Eq.1 & Eq.2

       α     y(1+λ)
2α - y.― = - ――――――
       β      1-λ

             α
2α.(1-λ) - y.―.(1-λ) = - y(1+λ)
             β

                         α
2α.(1-λ)  = - y(1+λ) - y.―.(1-λ)
                         β

                ┌               ┐
                │        α      │
2α.(1-λ)  = - y.│(1+λ) - ―.(1-λ)│
                │        β      │
                └               ┘

          2α.(1-λ) 
y = - ―――――――――――――――――
      ┌               ┐
      │        α      │ 
      │(1+λ) - ―.(1-λ)│
      │        β      │
      └               ┘

      2αβ.(λ-1)
y = ―――――――――――――
    α(λ-1)+β(λ+1)

-----------------------------
Now after doing the same for x...

      2αβ.(λ+1)
x = ―――――――――――――
    α(λ-1)+β(λ+1)

-----------------------------

So every pair...
            ┌                               ┐
            │   2αβ.(λ+1)       2αβ.(λ+1)   │
[ x , y ] = │ ――――――――――――― , ――――――――――――― │
            │ α(λ-1)+β(λ+1)   α(λ-1)+β(λ+1) │
            └                               ┘
 ...with α ≠ 0; β ≠ 0; x ≠ y

is a solution of the system in question.