Square-Root Equations

The equations are of the form

$\begin{aligned} \frac {A+B}{A-B} & = \dfrac {C+D}{C-D} \\ (A+B)(C-D) & = (A-B)(C+D) \\ AC-AD+BC-BD & = AC+AD-BC-BD \\ 2 BC & = 2AD \\ BC & = AD \\ B^2C^2 & = A^2D^2 \end{aligned}$

Then equation 1 becomes:

$\begin{aligned} (8x+y)(8x-y) & = (x+2)(5y+1) & ...(1a) \end{aligned}$

Equation 2:

$\begin{aligned} (3x-y)(4x+y) & = (x+y)(4x-y) \\ 12x^2 -xy - y^2 = 4x^2 + 3xy - y^2 \\ 8x^2-4xy & = 0 \\ 4x(2x-y) & = 0 & \small \color{#3D99F6} \text{Given that }x , y > 0 \\ \implies y & = 2x \quad ...(2a) \end{aligned}$

Putting $y=2x$ in equation 1a:

$\begin{aligned} (10x)(6x) & = (x+2)(10x+1) \\ 60x^2 & = 10x^2+21x +2 \\ 50x^2 - 21x - 2 & = 0 \\(25x+2)(2x-1) & = 0 \\ \implies x & = \frac 12 & \small \color{#3D99F6} \text{Since }x > 0 \\ y & = 2x = 1 \end{aligned}$

Therefore, $x+y = \dfrac 12 + 1 = \boxed{1.5}$ .