Question

find the solutions to the systems on nonlinear equations y=3x x^2+y^2=40

Answer 1

\[\begin{cases}y=3x\\x^2+y^2=40\end{cases}\]
Plug the first equation into the second:
\[x^2+(3x)^2=40\\
x^2+9x^2=40\\
10x^2=40\\
x^2=4\\
x=\pm2\]
Since \(y=3x\), that means \(y=3\times2=6\) and \(y=3\times(-2)=-6\).

Your solutions are then \((2,6)\) and \((-2,-6)\).