so these two lines intersect at x = 3 hmmmm what's the y-coordinate of the second line at that point?
[tex]\bf 2x-3y=6\implies \stackrel{\textit{using x = 3}}{2(3)-3y=6}\implies 6-3y=6\implies -3y=0\implies y=0[/tex]
so the y-coordinate of that second line is 0, so the point there is (3, 0) then.
so the first line is, a line that has a slope of 2 and passes through (3, 0).
[tex]\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{0})~\hspace{10em} slope = m\implies 2 \\\\\\ \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-0=2(x-3)\implies y=2x-6[/tex]
