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 $$\small{\text{$(\ x_2 = 3 \quad y_2 = 10 \ )$}}$$