Question

Solve the following system of equations.x² + y² = 20x² + 5y²=48

Answer 1

Given the system of equations:

`\begin{cases}x^2+y^2={20...(1)} \\ x^2+5y^2={48...(2)}\end{cases}`

We subtract equation (1) from equation (2):

`\begin{gathered} x^2+5y^2-x^2-y^2=48-20 \\ \\ 4y^2=28 \\ \\ y^2=7 \\ \\ \Rightarrow y=\pm√(7) \end{gathered}`

We use this value to find the corresponding x-values. Using (1):

`\begin{gathered} x^2+(\pm√(7))^2=20 \\ \\ x^2+7=20 \\ \\ x^2=13 \\ \\ \Rightarrow x=\pm√(13) \end{gathered}`

Finally, the solutions are:

`\begin{gathered} (x_1,y_1)=(-√(13),-√(7)) \\ \\ (x_2,y_2)=(-√(13),√(7)) \\ \\ (x_3,y_3)=(√(13),-√(7)) \\ \\ (x_4,y_4)=(√(13),√(7)) \end{gathered}`