Answer
$$\eqalign{
& \left( {\bf{a}} \right)L\left( {x,y} \right) = - 4x + 6y + 5 \cr
& \left( {\bf{b}} \right)L\left( {2.95,2.05} \right) = 5.5 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{{x + y}}{{x - y}};\,\,\,\,\left( {3,2} \right)\,\,\,{\text{ and }}\,\,\,\left( {2.95,2.05} \right) \cr
& f\left( {3,2} \right) = \frac{{3 + 2}}{{3 - 2}} \cr
& f\left( {3,2} \right) = 5 \cr
& {\text{The partial derivatives are}} \cr
& {f_x}\left( {x,y} \right) = - \frac{{2y}}{{{{\left( {x - y} \right)}^2}}} \cr
& {f_y}\left( {x,y} \right) = \frac{{2x}}{{{{\left( {x - y} \right)}^2}}} \cr
& {\text{Evaluate the partial derivatives at }}\left( {3,2} \right) \cr
& {f_x}\left( {3,2} \right) = - \frac{{2\left( 2 \right)}}{{{{\left( {3 - 2} \right)}^2}}} = - 4 \cr
& {f_y}\left( {3,2} \right) = \frac{{2\left( 3 \right)}}{{{{\left( {3 - 2} \right)}^2}}} = 6 \cr
& \cr
& \left( a \right) \cr
& {\text{The linear approximation to the function at }}\,\left( {3,2,5} \right){\text{ is}} \cr
& L\left( {x,y} \right) = {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_y}\left( {a,b} \right)\left( {y - b} \right) + f\left( {a,b} \right) \cr
& L\left( {x,y} \right) = - 4\left( {x - 3} \right) + 6\left( {y - 2} \right) + 5 \cr
& L\left( {x,y} \right) = - 4x + 12 + 6y - 12 + 5 \cr
& L\left( {x,y} \right) = - 4x + 6y + 5 \cr
& \cr
& \left( b \right){\text{Estimate }}\,\left( {2.95,2.05} \right) \cr
& L\left( {2.95,2.05} \right) = - 4\left( {2.95} \right) + 6\left( {2.05} \right) + 5 \cr
& L\left( {2.95,2.05} \right) = 5.5 \cr} $$
