Chapter 9 - Systems and Matrices - 9.5 Nonlinear Systems of Equations - 9.5 Exercises - Page 904: 30

$$\left\{ {\left( { - 2, - 1} \right),\left( { - 2,1} \right),\left( {2, - 1} \right),\left( {2,1} \right)} \right\}$$

$$\eqalign{ & 3{x^2} + 5{y^2} = 17\,\,\,\,\,\left( {\bf{1}} \right) \cr & \,2{x^2} - 3{y^2}\,\, = 5\,\,\,\,\,\,\,\left( {\bf{2}} \right) \cr & {\text{Multiply the equation }}\left( {\bf{1}} \right){\text{ by 3 and the equation }}\left( {\bf{2}} \right){\text{ by 5 and}} \cr & {\text{add both equations to eliminate }}{y^2}. \cr & 9{x^2} + 15{y^2} = 51\, \cr & \underline {10{x^2} - 15{y^2}\,\, = 25} \cr & 19{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 76 \cr & \cr & {\text{Solve the quadratic equation 19}}{x^2} = 76 \cr & {x^2} = 4 \cr & x = \pm 2 \cr & {x_1} = - 2,\,\,\,{x_2} = 2 \cr & \cr & {\text{From the equation }}\left( {\bf{1}} \right){\text{ we have that}} \cr & \,3{x^2} + 5{y^2} = 17 \cr & {y^2} = \frac{{17 - 3{x^2}}}{5} \cr & \cr & {\text{Substitute }}{x_1} = - 2{\text{ into the equation }}{y^2} = \frac{{17 - 3{x^2}}}{5} \cr & {y^2} = \frac{{17 - 3{x^2}}}{5} \cr & {y^2} = \frac{{17 - 3{{\left( { - 2} \right)}^2}}}{5} \cr & y = \pm 1 \cr & {\text{The first and second solutions are }}\left( { - 2, - 1} \right){\text{ and }}\left( { - 2,1} \right) \cr & \cr & {\text{Substitute }}{x_2} = 2{\text{ into the equation }}{y^2} = \frac{{17 - 3{x^2}}}{5} \cr & {y^2} = \frac{{17 - 3{x^2}}}{5} \cr & {y^2} = \frac{{17 - 3{{\left( 2 \right)}^2}}}{5} \cr & y = \pm 1 \cr & {\text{The third and fourth solutions are }}\left( {2, - 1} \right){\text{ and }}\left( {2,1} \right) \cr & \cr & {\text{Therefore, the solution set of the system is}} \cr & \left\{ {\left( { - 2, - 1} \right),\left( { - 2,1} \right),\left( {2, - 1} \right),\left( {2,1} \right)} \right\} \cr} $$