[tex]\bf y=\stackrel{\stackrel{slope}{\downarrow }}{-\cfrac{2}{3}}x-2\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]
now, since parallel lines have the same slopes, a parallel line to the one above will also have a slope of -2/3, well then, we're really looking for the equation of a line whose slope is -2/3 and passes through (3,3)
[tex]\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{3})~\hspace{10em} slope = m\implies -\cfrac{2}{3} \\\\\\ \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-3=-\cfrac{2}{3}(x-3) \\\\\\ y-3=-\cfrac{2}{3}x+2\implies y=-\cfrac{2}{3}x+5[/tex]
