Find equations of the tangent lines to the curve y = ( x − 1) / (x + 1) that are parallel to the line x − 2y = 5.

Answer:[tex]y = \frac{1}{2}x+\frac{7}{2}[/tex][tex]y = \frac{1}{2}x-\frac{1}{2}[/tex]Step-by-step explanation:The first step is obtain the derivative of the function y:[tex]\frac{dy}{dx}=\frac{(x+1)*(1)-(x-1)*1}{(x+1)^{2} }=\frac{2}{(x+1)^{2} }[/tex](Remember that the slope of the tangent line to any curve is given by its derivative).Secondly, in order to obtain the equations of the tangent lines parallel to x - 2y = 5 we need to obtain the slope of this line, remember that the line equation is given by:[tex]y = mx + b[/tex]m : slopeb: y-interceptThen, the slope of the line[tex]y = \frac{1}{2}x-\frac{5}{2}[/tex]is m = 1/2.Then, the derivative of the function y must be 1/2:[tex]y= \frac{2}{(x+1)^{2}}=\frac{1}{2}\\(x+1)^{2}=4\\x^{2}+2x-3=0\\ (x+3)(x-1)=0\\x = -3\\x=1[/tex]When x = -3, y =(-3-1)/(-3+1)= 2, then one of the tangent lines must pass through the point (-3,2). When x = 1, y = (1-1)(1+1)= 0, then the other tangent line must pass through the point (1, 0).Finally, the point-slope equation of a line is given by:[tex]y- y_{1}  = m(x - x_{1})[/tex]Substituting our previous results we have the equations of the tangent lines:[tex]y - 2 = \frac{1}{2}(x - (-3))\\ y = \frac{1}{2}x+\frac{7}{2}[/tex][tex]y - 0 = \frac{1}{2}(x- 1) \\y = \frac{1}{2}x-\frac{1}{2}[/tex]