Unknowns are Unknowns! (Part 2)

$\begin{aligned} \frac {x-y}{x+y} & = \frac ab & \small \color{#3D99F6} \text{Squaring both sides} \\ \frac {x^2-2xy+y^2}{x^2+2xy+y^2} & = \frac {a^2}{b^2} & \small \color{#3D99F6} \text{Multiplying up and down of LHS by }2 \\ \frac {{\color{#3D99F6}2x^2}-4xy+{\color{#3D99F6}2y^2}}{{\color{#3D99F6}2x^2}+4xy+{\color{#3D99F6}2y^2}} & = \frac {a^2}{b^2} & \small \color{#3D99F6} \text{Note that } 2x^2+2y^2 = 5xy \\ \frac {{\color{#3D99F6}5xy}-4xy}{{\color{#3D99F6}5xy}+4xy} & = \frac {a^2}{b^2} \\ \implies \frac {a^2}{b^2} & = \frac 19 \\ \frac ab & = \frac 13 \end{aligned}$

$\implies a \times b - a + b \div a = 1 \times 3 - 1 + 3 \div 1 = \boxed{5}$

If $y = 0$ then from the first equation $x = 0$ , which would make $(x - y)/(x + y)$ indeterminate. So with $y \ne 0$ we can divide the first equation through by $y^{2}$ to find that

$2\left(\dfrac{x}{y}\right)^{2} + 2 = 5\left(\dfrac{x}{y}\right) \Longrightarrow 2\left(\dfrac{x}{y}\right)^{2} - 5\left(\dfrac{x}{y}\right) + 2 = 0 \Longrightarrow \left(2\dfrac{x}{y} - 1\right)\left(\dfrac{x}{y} - 2\right) = 0$ .

So either $\dfrac{x}{y} = \dfrac{1}{2}$ or $\dfrac{x}{y} = 2$ . Now the second equation can be written as $\dfrac{\dfrac{x}{y} - 1}{\dfrac{x}{y} + 1} = \dfrac{a}{b}$ ,

so if we want $a,b$ to be positive we need to choose $\dfrac{x}{y} = 2$ , in which case $a = 1, b = 3$ .

Thus $a \times b - a + b \div a = 1 \times 3 - 1 + 3 \div 1 = \boxed{5}$ .

$\begin{aligned} 2x^2 + 2y^2 & = 5xy \\ 2(x^2 + y^2) & = 5xy \\ x^2 + y^2 & = \dfrac{5}{2}xy \end{aligned}$

Note that $(x-y)^2 = x^2 + y^2 -2xy \rightarrow (x-y)^2 = \dfrac{5}{2}xy - 2xy = \dfrac{1}{2}xy$ .

Also, $(x+y)^2 = x^2 + y^2 + 2xy \rightarrow (x+y)^2 = \dfrac{5}{2}xy + 2xy = \dfrac{9}{2}xy$ .

Now,

$\begin{aligned} \dfrac{x-y}{x+y} = \sqrt{ \dfrac{(x-y)^2}{(x+y)^2}} & = \sqrt{ \dfrac{ \dfrac{1}{2}xy}{ \dfrac{9}{2}xy}} \\ & = \sqrt{ \dfrac{1}{9}} = \dfrac{1}{3} = \dfrac{a}{b} \end{aligned}$

Hence $a \times b - a + b : a = 3 - 1 + 3 = 5$ .

You could just find a pair of (x,y) that works. I found that x=4, y=2.