Chapter 9 - Systems and Matrices - 9.5 Nonlinear Systems of Equations - 9.5 Exercises - Page 904: 44

$$\left\{ {\left( {4i, - 2i} \right),\left( { - 4i,2i} \right),\left( {2,4} \right),\left( { - 2, - 4} \right)} \right\}$$

$$\eqalign{ & \,{x^2} + 3xy - {y^2} = 12\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \,\,\,\,\,{x^2} - {y^2} = - 12\,\,\,\,\,\,\,\,\,\,\,\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} + 3xy - {y^2} = 12 \cr & \underline { - {x^2}\,\,\,\,\,\,\,\,\,\,\,\,\, + {y^2} = 12\,\,\,\,\,\,\,\,\,\,\,\,} \cr & \,\,\,\,\,\,\,\,\,\,\,\,3xy\,\,\,\,\,\,\,\,\,\,\,\, = 24\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{3}} \right) \cr & \cr & {\text{Solve the equation }}\left( {\bf{3}} \right){\text{ for }}y \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = \frac{8}{x} \cr & \cr & {\text{Substitute }}\frac{8}{x}{\text{ for }}y{\text{ into the equation }}\left( {\bf{2}} \right) \cr & {x^2} - {\left( {\frac{8}{x}} \right)^2} = - 12\, \cr & {\text{Solve for }}x \cr & {x^2} - \frac{{64}}{{{x^2}}} = - 12\, \cr & {x^4} - 64 = - 12{x^2} \cr & {x^4} + 12{x^2} - 64 = 0 \cr & \left( {{x^2} + 16} \right)\left( {{x^2} - 4} \right) = 0 \cr & {x_1} = 4i,\,\,\,{x_2} = - 4i,\,\,\,\,{x_3} = 2,\,\,\,\,\,{x_4} = - 2 \cr & \cr & {\text{Substitute }}{x_1}{\text{ = }}4i{\text{ for }}x{\text{ into the equation }}y = \frac{8}{x} \cr & y = \frac{8}{{4i}} = - 2i \cr & {\text{The first solution is }}\left( {4i, - 2i} \right) \cr & \cr & {\text{Substitute }}{x_2}{\text{ = }} - 4i{\text{ for }}x{\text{ into the equation }}y = \frac{8}{x} \cr & y = \frac{8}{{ - 4i}} = 2i \cr & {\text{The second solution is }}\left( { - 4i,2i} \right) \cr & \cr & {\text{Substitute }}{x_3}{\text{ = }}2{\text{ for }}x{\text{ into the equation }}y = \frac{{15}}{x} \cr & y = \frac{8}{2} = 4 \cr & {\text{The third solution is }}\left( {2,4} \right) \cr & \cr & {\text{Substitute }}{x_4}{\text{ = }} - 2{\text{ for }}x{\text{ into the equation }}y = \frac{{15}}{x} \cr & y = \frac{8}{{ - 2}} = - 4 \cr & {\text{The third solution is }}\left( { - 2, - 4} \right) \cr & \cr & {\text{Therefore, the solution set of the system is}} \cr & \left\{ {\left( {4i, - 2i} \right),\left( { - 4i,2i} \right),\left( {2,4} \right),\left( { - 2, - 4} \right)} \right\} \cr} $$