solve the following nonlinear system algebraically. be sure to check for non-real solutions.

Use substitution method to and solve in terms of x using the first equation
[tex]\begin{gathered} x^2+y^2=8 \\ x^2=8-y^2 \end{gathered}[/tex]Next, substitute it to the second equation
[tex]\begin{gathered} 2x^2+4y^2=34 \\ 2(8-y^2)+4y^2=34 \\ 16-2y^2+4y^2=34 \\ 2y^2=34-16 \\ 2y^2=18 \\ \frac{2y^2}{2}=\frac{18}{2} \\ y^2=9 \end{gathered}[/tex]Then, substitute it back to first equation and solve for x
[tex]\begin{gathered} x^2+y^2=8 \\ x^2+9=8 \\ x^2=8-9 \\ x^2=-1 \end{gathered}[/tex]Now we have the following solutions
[tex]x^2=-1\text{ and }y^2=9[/tex]Get the square root of both x's and y's to get the solution
[tex]\begin{gathered} x^2=-1 \\ \sqrt{x^2}=\sqrt{-1} \\ x=\pm i \\ x=i\text{ and }x=-i \\ \\ y^{2}=9 \\ \sqrt{y^2}=\sqrt{9} \\ y=\operatorname{\pm}3 \\ y=3\text{ and }y=-3 \end{gathered}[/tex]Getting the combination of ordered pairs we have the following solutions
[tex]\begin{gathered} (i,3) \\ (i,-3) \\ (-i,3) \\ (-i,-3) \end{gathered}[/tex]