BEUse substitution to solve.f2x2 = 5 + y4y = -20 + 8x2Solve the first equation for y and substitute it into the second equation. The resulting equati4y = 16x2 - 608x2 - 20 = -20 + 8x22x2 = 5 + 2x2 - 5

Given that:
[tex]\begin{gathered} 2x^2=5+y \\ 4y=-20+8x^2 \end{gathered}[/tex]From the first equation,
[tex]y=5-2x^2[/tex]Substitute the obtained value of y into the second equation.
[tex]\begin{gathered} 4(5-2x^2)=-20+8x^2 \\ 20-8x^2=-20+8x^2 \end{gathered}[/tex]So, it has infintely many solutions.
Solutions are of the form:
[tex](x,y)=(x,5-2x^2[/tex]where x is any real number.