Solve the system

y=3x+1 ( line )

y=x^2+1 ( parabola )

$$\small{\text{
$
\begin{array}{rcl}
y_{parabola} &=& y_{line} \\
x^2+1 &=& 3x+1\\
x^2+\not{1} &=& 3x+\not{1}\\
x^2-3x &=& 3x \\
x^2-3x &=& 0 \\
\underbrace{x}_{x_1=0}\cdot\underbrace{(x-3)}_{=0} &=& 0 \\
x_2-3 &=& 0 \\
x_2 &=& 3
\end{array}
$
}}$$

$$\small{\text{
$
\begin{array}{rcl}
\\y_1 &=& 3\cdot x_1 + 1 \\
y_1 &=& 3 \cdot 0 + 1 \\
y_1 &=& 1\\\\
y_2 &=& 3\cdot x_2 + 1 \\
y_2 &=& 3 \cdot 3 + 1 \\
y_2 &=& 10
\end{array}
$
}}$$

So we have: $$\small{\text{$(\ x_1 = 0 \quad y_1 = 1 \ )$}}$$ and $$\small{\text{$(\ x_2 = 3 \quad y_2 = 10 \ )$}}$$