Respuesta :
First, let's solve the second equation for y:
[tex]\begin{gathered} y=x+5y \\ 5y-y=-x \\ 4y=-x \\ y=-\frac{x}{4} \end{gathered}[/tex]Now, let's use this value of y in the first equation:
[tex]\begin{gathered} y=x^2+3 \\ -\frac{x}{4}=x^2+3 \\ x^2+\frac{x}{4}+3=0 \\ 4x^2+x+12=0 \end{gathered}[/tex]Let's solve this quadratic equation using the quadratic formula:
[tex]\begin{gathered} a=4,b=1,c=12 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x_1=\frac{-1+\sqrt[]{1-192}}{8}=\frac{-1+\sqrt[]{-191}}{8} \\ x_2=\frac{-1-\sqrt[]{-191}}{8} \end{gathered}[/tex]Now, calculating y, we have:
[tex]\begin{gathered} y_1=-\frac{x_1}{4}=\frac{1-\sqrt[]{-191}}{32} \\ y_2=\frac{-x_2}{4}=\frac{1+\sqrt[]{-191}}{32} \end{gathered}[/tex]