Check the picture below.
to get the equation of any straight line, we simply need any two points off of it, so let's use those two from the picture.
[tex](\stackrel{x_1}{-3}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{9}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{9}-\stackrel{y1}{(-6)}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{(-3)}}} \implies \cfrac{9 +6}{2 +3} \implies \cfrac{ 15 }{ 5 }\implies 3[/tex]
[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{3}(x-\stackrel{x_1}{(-3)}) \\\\\\ y+6=3(x+3)\implies y+6=3x+9\implies y=3x+3[/tex]
