Chapter 9 - Systems and Matrices - 9.5 Nonlinear Systems of Equations - 9.5 Exercises - Page 904: 29

$$\left\{ {\left( { - 1, - 1} \right),\left( { - 1,1} \right),\left( {1, - 1} \right),\left( {1,1} \right)} \right\}$$

$$\eqalign{ & \,2{x^2} + 3{y^2} = 5\,\,\,\,\,\,\left( {\bf{1}} \right) \cr & 3{x^2} - 4{y^2}\,\, = - 1\,\,\,\,\left( {\bf{2}} \right) \cr & {\text{Multiply the equation }}\left( {\bf{1}} \right){\text{ by }}4{\text{ and the equation }}\left( {\bf{2}} \right){\text{ by 3 and}} \cr & {\text{add both equations to eliminate }}{y^2}. \cr & 8{x^2} + 12{y^2} = 20 \cr & \underline {9{x^2} - 12{y^2}\,\, = - 3} \cr & 17{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 17 \cr & \cr & {\text{Solve the quadratic equation 17}}{x^2} = 17 \cr & {x^2} = 1 \cr & x = \pm 1 \cr & {x_1} = - 1,\,\,\,{x_2} = 1 \cr & \cr & {\text{From the equation }}\left( {\bf{1}} \right){\text{ we have that}} \cr & 2{x^2} + 3{y^2} = 5 \cr & {y^2} = \frac{{5 - 2{x^2}}}{3} \cr & \cr & {\text{Substitute }}{x_1} = - 1{\text{ into the equation }}{y^2} = \frac{{5 - 2{x^2}}}{3} \cr & {y^2} = \frac{{5 - 2{{\left( { - 1} \right)}^2}}}{3} \cr & {y^2} = 1 \cr & y = \pm 1 \cr & {\text{The first and second solutions are }}\left( { - 1, - 1} \right){\text{ and }}\left( { - 1,1} \right) \cr & \cr & {\text{Substitute }}{x_2} = 1{\text{ into the equation }}{y^2} = \frac{{5 - 2{x^2}}}{3} \cr & {y^2} = \frac{{5 - 2{{\left( 1 \right)}^2}}}{3} \cr & {y^2} = 1 \cr & y = \pm 1 \cr & {\text{The third and fourth solutions are }}\left( {1, - 1} \right){\text{ and }}\left( {1,1} \right) \cr & \cr & {\text{Therefore, the solution set of the system is}} \cr & \left\{ {\left( { - 1, - 1} \right),\left( { - 1,1} \right),\left( {1, - 1} \right),\left( {1,1} \right)} \right\} \cr} $$