Chapter 9 - Systems and Matrices - 9.5 Nonlinear Systems of Equations - 9.5 Exercises - Page 904: 27

$$\left\{ {\left( { - i,\sqrt 6 } \right),\left( { - i, - \sqrt 6 } \right),\left( {i,\sqrt 6 } \right),\left( {i, - \sqrt 6 } \right)} \right\}$$

$$\eqalign{ & \,\,\,\,\,\,3{x^2} + {y^2} = 3\,\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & \,\,\,4{x^2} + 5{y^2} = 26\,\,\,\,\left( {\bf{2}} \right) \cr & {\text{Multiply the equation }}\left( {\bf{1}} \right){\text{ by }} - 5{\text{ and add both equations to}} \cr & {\text{eliminate }}{y^2} \cr & - 15{x^2} - 5{y^2} = - 15 \cr & \underline {\,\,\,\,\,4{x^2} + 5{y^2} = 26} \cr & - 11{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 11 \cr & \cr & {\text{Solve the quadratic equation }} - {x^2} = 11 \cr & {x^2} = - 1 \cr & x = \pm i \cr & {x_1} = - i,\,\,\,{x_2} = i \cr & \cr & {\text{From the equation }}\left( {\bf{1}} \right){\text{ we have that}} \cr & 3{x^2} + {y^2} = 3 \cr & {y^2} = 3 - 3{x^2} \cr & \cr & {\text{Substitute }}{x_1} = - i{\text{ into the equation }}{y^2} = 3 - 3{x^2} \cr & {y^2} = 3 - 3{\left( { - i} \right)^2} \cr & {y^2} = 6 \cr & y = \pm \sqrt 6 \cr & {\text{The first and second solutions are }}\left( { - i,\sqrt 6 } \right){\text{ and }}\left( { - i, - \sqrt 6 } \right) \cr & \cr & {\text{Substitute }}{x_2} = i{\text{ into the equation }}{y^2} = 3 - 3{x^2} \cr & {y^2} = 3 - 3{\left( i \right)^2} \cr & {y^2} = 6 \cr & y = \pm \sqrt 6 \cr & {\text{The third and fourth solutions are }}\left( {i,\sqrt 6 } \right){\text{ and }}\left( {i, - \sqrt 6 } \right) \cr & \cr & {\text{Therefore, the solution set of the system is}} \cr & \left\{ {\left( { - i,\sqrt 6 } \right),\left( { - i, - \sqrt 6 } \right),\left( {i,\sqrt 6 } \right),\left( {i, - \sqrt 6 } \right)} \right\} \cr} $$