Respuesta :
Given the system of equations:
[tex]\begin{cases}x^2+y^2={20...(1)} \\ x^2+5y^2={48...(2)}\end{cases}[/tex]We subtract equation (1) from equation (2):
[tex]\begin{gathered} x^2+5y^2-x^2-y^2=48-20 \\ \\ 4y^2=28 \\ \\ y^2=7 \\ \\ \Rightarrow y=\pm\sqrt{7} \end{gathered}[/tex]We use this value to find the corresponding x-values. Using (1):
[tex]\begin{gathered} x^2+(\pm\sqrt{7})^2=20 \\ \\ x^2+7=20 \\ \\ x^2=13 \\ \\ \Rightarrow x=\pm\sqrt{13} \end{gathered}[/tex]Finally, the solutions are:
[tex]\begin{gathered} (x_1,y_1)=(-\sqrt{13},-\sqrt{7}) \\ \\ (x_2,y_2)=(-\sqrt{13},\sqrt{7}) \\ \\ (x_3,y_3)=(\sqrt{13},-\sqrt{7}) \\ \\ (x_4,y_4)=(\sqrt{13},\sqrt{7}) \end{gathered}[/tex]