#### Answer

$ y-2=\displaystyle \frac{2}{3}(x+2) \qquad$ ... point-slope form
$ y=\displaystyle \frac{2}{3}x+\frac{10}{3} \qquad$ ... slope-intercept form

#### Work Step by Step

The line ...
$ 2x-3y=7\qquad$... solve for y
$-3y=-2x+7 \qquad$... $/\div(-3)$
$y=\displaystyle \frac{2}{3}x-\frac{7}{3}$
.. is now in slope-intercept form. Its slope is $m=\displaystyle \frac{2}{3}.$
A line parallel to it has the same slope, $m=\displaystyle \frac{2}{3}$
and passes through $(x_{1},y_{1})=(-2,2)$
We write the point-slope equation:
$y-y_{1}=m(x-x_{1})$
$ y-2=\displaystyle \frac{2}{3}(x+2) \qquad$ ... point-slope form
Simplify to slope-intercept form (solve for y)
$ y-2=\displaystyle \frac{2}{3}x+\frac{4}{3} \qquad$ ...add $2$
$ y=\displaystyle \frac{2}{3}x+\frac{10}{3} \qquad$ ... is the slope-intercept form