Substituting the second equation in the first one we get:
[tex]x^2+(3x+1)^2=5.[/tex]
Simplifying the above equation we get:
[tex]\begin{gathered} x^2+(3x)^2+2*(3x)(1)+1^2=5, \\ x^2+9x^2+6x+1=5, \\ 10x^2+6x+1=5. \end{gathered}[/tex]
Subtracting 5 from the above equation we get:
[tex]\begin{gathered} 10x^2+6x+1-5=5-5, \\ 10x^2+6x-4=0. \end{gathered}[/tex]
Now, notice that:
[tex]10x^2+6x-4=2(5x^2+3x-2)=2(5x-2)(x+1).[/tex]
Therefore:
[tex]10x^2+6x-4=0\text{ if and only if }x=\frac{2}{5}\text{ or }x=-1.[/tex]
If x=2/5:
[tex]y=3*\frac{2}{5}+1=\frac{6}{5}+1=\frac{11}{5}.[/tex]
Therefore
[tex](\frac{2}{5},\frac{11}{5})[/tex]
is a solution to the given system of equations.
If x=-1:
[tex]y=3*(-1)+1=-3+1=-2.[/tex]
Therefore (-1,-2) is a solution to the given system of equations.
Answer: Third option.
