Chapter 9 - Systems and Matrices - 9.5 Nonlinear Systems of Equations - 9.5 Exercises - Page 904: 33

$$\left\{ {\left( {\sqrt 6 ,0} \right),\left( { - \sqrt 6 ,0} \right)} \right\}$$

$$\eqalign{ & \,2{x^2} - 3{y^2} = 12\,\,\,\,\,\left( {\bf{1}} \right) \cr & \,6{x^2} + 5{y^2} = 36\,\,\,\,\left( {\bf{2}} \right) \cr & {\text{Multiply the equation }}\left( {\bf{1}} \right){\text{ by}} - 5{\text{ and the equation }}\left( {\bf{2}} \right){\text{ by}}\,3 \cr & {\text{then, add both equations}}{\text{.}} \cr & - 10{x^2} + 15{y^2} = - 60 \cr & \underline {\,\,\,18{x^2} + 15{y^2} = 108} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8{x^2}\, = 48 \cr & {\text{Solve the quadratic equation }}8{x^2}\, = 48 \cr & 8{x^2}\, = 48 \cr & {x^2}\, = 6 \cr & {x_1} = - \sqrt 6 ,\,\,\,{x_2} = \sqrt 6 \cr & \cr & {\text{From the equation }}\left( {\bf{2}} \right){\text{ we have that}} \cr & 6{x^2} + 5{y^2} = 36\, \cr & {y^2} = \frac{{36 - 6{x^2}}}{5} \cr & \cr & {\text{Substitute }}{x_1} = - \sqrt 6 {\text{ into the equation }}{y^2} = \frac{{36 - 6{x^2}}}{5} \cr & {y^2} = \frac{{36 - 6{{\left( { - \sqrt 6 } \right)}^2}}}{5} \cr & {y^2} = 0 \cr & y = 0 \cr & {\text{The first solution is }}\left( { - \sqrt 6 ,0} \right) \cr & \cr & {\text{Substitute }}{x_2} = \sqrt 6 {\text{ into the equation }}{y^2} = \frac{{36 - 6{x^2}}}{5} \cr & {y^2} = \frac{{36 - 6{{\left( {\sqrt 6 } \right)}^2}}}{5} \cr & {y^2} = 0 \cr & y = 0 \cr & {\text{The second solution is }}\left( {\sqrt 6 ,0} \right) \cr & \cr & {\text{Therefore, the solution set of the system is}} \cr & \left\{ {\left( {\sqrt 6 ,0} \right),\left( { - \sqrt 6 ,0} \right)} \right\} \cr} $$