Solution to Practice Problem #4


\[\begin{array}{c} \eqalign{ & {{9x + 12} \over {{x^2} – 16}} = {R \over {x + 4}} + {S \over {x – 4}} \cr & \cr & {{9x + 12} \over {{x^2} – 16}} = {R \over {x + 4}}\left( {{{x – 4} \over {x – 4}}} \right) + {S \over {x – 4}}\left( {{{x + 4} \over {x + 4}}} \right) \cr & \cr & {{9x + 12} \over {{x^2} – 16}} = {{R\,x – 4\,R} \over {{x^2} – 16}} + {{S\,x + 4\,S} \over {{x^2} – 16}} \cr & \cr & {{9x + 12} \over {{x^2} – 16}} = {{R\,x – 4\,R + S\,x + 4\,S} \over {{x^2} – 16}} \cr & \cr & {{9x + 12} \over {{x^2} – 16}} = {{\left( {R\,x + S\,x} \right) + \left( { – 4\,R + 4\,S} \right)} \over {{x^2} – 16}} \cr & \cr & {{9x + 12} \over {{x^2} – 16}} = {{\left( {R\, + S\,} \right)x + \left( { – 4\,R + 4\,S} \right)} \over {{x^2} – 16}} \cr & \cr & \,\,\,\,\,\,\,R\,\,\, + \,\,\,\,\,S\, = \,\,\,9 \cr & – 4\,R\,\,\, + \,\,4\,S = 12 \cr & \cr & \,\,\,\,4R\,\,\, + \,\,\,\,4\,S\, = \,\,36 \cr & – 4\,R\,\,\, + \,\,\,\,4\,S = \,\,12 \cr & \cr & 8\,S = 48 \cr & S = 6 \cr & \cr & R = 3 \cr & \cr & R + S = 9 \cr} \end{array}\]