Answer
$$\eqalign{
& y = - \frac{1}{2}x + \frac{7}{2} \cr
& y = - \frac{1}{2}x - \frac{1}{2} \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \frac{{x + 1}}{{x - 1}} \cr
& {\text{Find }}f'\left( x \right) \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{x + 1}}{{x - 1}}} \right] \cr
& {\text{By the quotient rule}} \cr
& f'\left( x \right) = \frac{{\left( {x - 1} \right)\left( 1 \right) - \left( {x + 1} \right)\left( 1 \right)}}{{{{\left( {x - 1} \right)}^2}}} \cr
& f'\left( x \right) = \frac{{x - 1 - x - 1}}{{{{\left( {x - 1} \right)}^2}}} \cr
& f'\left( x \right) = - \frac{2}{{{{\left( {x - 1} \right)}^2}}} \cr
& 2y + x = 6 \to m = - \frac{1}{2} \cr
& {\text{Find the point }}x{\text{ at which the slope is }}m = - \frac{1}{2} \cr
& - \frac{2}{{{{\left( {x - 1} \right)}^2}}} = - \frac{1}{2} \cr
& \frac{{{{\left( {x - 1} \right)}^2}}}{2} = 2 \cr
& {\left( {x - 1} \right)^2} = 4 \cr
& x - 1 = \pm 2 \cr
& {x_1} = - 1,{\text{ }}{x_2} = 3 \cr
& f\left( { - 1} \right) = 0,{\text{ }}f\left( 3 \right) = \frac{{3 + 1}}{{3 - 1}} = 2 \cr
& {\text{We obtain the points }}\left( { - 1,0} \right){\text{ and }}\left( {3,2} \right) \cr
& {\text{For }}\left( { - 1,0} \right) \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 0 = - \frac{1}{2}\left( {x + 1} \right) \to y = - \frac{1}{2}x - \frac{1}{2} \cr
& {\text{For }}\left( {3,2} \right) \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 2 = - \frac{1}{2}\left( {x - 3} \right) \to y = - \frac{1}{2}x + \frac{7}{2} \cr
& {\text{Graph}} \cr} $$
