Chapter 9 - Systems and Matrices - 9.5 Nonlinear Systems of Equations - 9.5 Exercises - Page 904: 34

$$\left\{ {\left( {\sqrt 5 ,0} \right),\left( { - \sqrt 5 ,0} \right)} \right\}$$

$$\eqalign{ & \,\,\,5{x^2} - 2{y^2} = 25\,\,\,\,\,\left( {\bf{1}} \right) \cr & \,\,\,10{x^2} + {y^2} = 50\,\,\,\,\left( {\bf{2}} \right) \cr & {\text{Multiply the equation }}\left( {\bf{2}} \right){\text{ by }}2{\text{ and add both equations to}} \cr & {\text{eliminate }}{y^2} \cr & 5{x^2} - 2{y^2} = 25 \cr & \underline {10{x^2} + {y^2} = 50} \cr & \,15{x^2}\,\,\,\,\,\,\,\,\,\,\, = 75 \cr & {\text{Solve the quadratic equation }}15{x^2}\, = 75 \cr & 15{x^2}\, = 75 \cr & {x^2}\, = 5 \cr & {x_1} = - \sqrt 5 ,\,\,\,{x_2} = \sqrt 5 \cr & \cr & {\text{From the equation }}\left( {\bf{2}} \right){\text{ we have that}} \cr & 10{x^2} + {y^2} = 50 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{y^2} = 50 - 10{x^2} \cr & \cr & {\text{Substitute }}{x_1} = - \sqrt 5 {\text{ into the equation }}{y^2} = 50 - 10{x^2} \cr & {y^2} = 50 - 10{\left( { - \sqrt 5 } \right)^2} \cr & {y^2} = 0 \cr & y = 0 \cr & {\text{The first solution is }}\left( { - \sqrt 5 ,0} \right) \cr & \cr & {\text{Substitute }}{x_2} = \sqrt 6 {\text{ into the equation }}{y^2} = 50 - 10{x^2} \cr & {y^2} = 50 - 10{\left( {\sqrt 5 } \right)^2} \cr & {y^2} = 0 \cr & y = 0 \cr & {\text{The second solution is }}\left( {\sqrt 5 ,0} \right) \cr & \cr & {\text{Therefore, the solution set of the system is}} \cr & \left\{ {\left( {\sqrt 5 ,0} \right),\left( { - \sqrt 5 ,0} \right)} \right\} \cr} $$