Answer
There are two tangent lines satisfying the conditions:
$$x-2y+7=0$$ $$x-2y-1=0$$
Work Step by Step
Given curve, $y=\frac{x-1}{x+1}$. The tangent line to this curve is parallel to $x-2y=2$, which means that the slope of the tangent line is $\frac{1}{2}$ (the slope of the line it is parallel to).
Differentiating y with respect to x,
$$\frac{dy}{dx}=\frac{x+1-x+1}{(x+1)^2}=\frac{2}{(x+1)^2}$$
This slope must be equal to $1/2$, that is, the slope of the required tangent.
$$\frac{2}{(x+1)^2}=\frac12\implies x^2+1+2x=4$$$$x^2+2x-3=0\implies x=-3,1.$$
Thus, the tangent touches the curve y at $(-3,y(-3))≡(-3,2)$ & $(1,y(1))≡(1,0)$. Thus, there are two tangent lines. For first point,
$$y-2=\frac12(x+3)\implies2y-4=x+3$$$$x-2y+7=0$$For second point, $$y-0=\frac12(x-1)\implies2y=x-1$$$$x-2y-1=0.$$
