Chapter 9 - Systems and Matrices - 9.5 Nonlinear Systems of Equations - 9.5 Exercises - Page 904: 43

$$\left\{ {\left( { - 5i,3i} \right),\left( {5i, - 3i} \right),\left( {3,5} \right),\left( { - 3, - 5} \right)} \right\}$$

$$\eqalign{ & \,{x^2} + 2xy - {y^2} = 14\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \,\,\,\,\,{x^2} - {y^2} = - 16\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{2}} \right) \cr & \cr & {\text{Multiply the equation }}\left( {\bf{2}} \right){\text{ by }} - 1{\text{ and add both equations to }} \cr & {\text{eliminate }}{x^2}{\text{ and }}{y^2} \cr & {x^2} + 2xy - {y^2} = 14 \cr & \underline { - {x^2}\,\,\,\,\,\,\,\,\,\,\,\,\, + {y^2} = 16\,\,\,\,\,\,\,\,\,\,\,\,} \cr & 2xy\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 30\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{3}} \right) \cr & \cr & {\text{Solve the equation }}\left( {\bf{3}} \right){\text{ for }}y \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \frac{{15}}{x} \cr & \cr & {\text{Substitute }}\frac{{15}}{x}{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr & {x^2} - {\left( {\frac{{15}}{x}} \right)^2} = - 16 \cr & {\text{Solve for }}x \cr & {x^2} - \frac{{225}}{{{x^2}}} = - 16 \cr & {x^4} - 225 = - 16{x^2} \cr & {x^4} + 16{x^2} - 225 = 0 \cr & \left( {{x^2} + 25} \right)\left( {{x^2} - 9} \right) = 0 \cr & {x_1} = 5i,\,\,\,{x_2} = - 5i,\,\,\,\,{x_3} = 3,\,\,\,\,\,{x_4} = - 3 \cr & \cr & {\text{Substitute }}{x_1}{\text{ = }}5i{\text{ for }}x{\text{ into the equation }}y = \frac{{15}}{x} \cr & y = \frac{{15}}{{5i}} = - 3i \cr & {\text{The first solution is }}\left( {5i, - 3i} \right) \cr & \cr & {\text{Substitute }}{x_2}{\text{ = }} - 5i{\text{ for }}x{\text{ into the equation }}y = \frac{{15}}{x} \cr & y = \frac{{15}}{{ - 5i}} = 3i \cr & {\text{The second solution is }}\left( { - 5i,3i} \right) \cr & \cr & {\text{Substitute }}{x_3}{\text{ = }}3{\text{ for }}x{\text{ into the equation }}y = \frac{{15}}{x} \cr & y = \frac{{15}}{3} \cr & {\text{The third solution is }}\left( {3,5} \right) \cr & \cr & {\text{Substitute }}{x_4}{\text{ = }} - 3{\text{ for }}x{\text{ into the equation }}y = \frac{{15}}{x} \cr & y = \frac{{15}}{{ - 3}} = - 5 \cr & {\text{The third solution is }}\left( { - 3, - 5} \right) \cr & \cr & {\text{Therefore, the solution set of the system is}} \cr & \left\{ {\left( { - 5i,3i} \right),\left( {5i, - 3i} \right),\left( {3,5} \right),\left( { - 3, - 5} \right)} \right\} \cr} $$