keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above
[tex]y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x-7\qquad \impliedby \qquad \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]
so, we're really looking for the equation of a line whose slope is -2/3 and passes through (-6 , 3)
[tex](\stackrel{x_1}{-6}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{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-\stackrel{y_1}{3}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{(-6)})\implies y-3=-\cfrac{2}{3}(x+6) \\\\\\ y-3=-\cfrac{2}{3}x-4\implies y=-\cfrac{2}{3}-1[/tex]
The equation for the line parallel to y =-2/3x – 7 that contains (-6, 3) is y - 3 = -2/3(x+6)
The equation of a line in point slope form is expressed as:
y - y1 = m(x-x1)
m is the slope
(x1, y1) is any point on the line
GIven the equation y =-2/3x – 7, the slope of the line is -2/3
Substitute m = -2/3 and (-6, 3) into the formula
y - 3 = -2/3(x-(-6)
y - 3 = -2/3(x+6)
Hence the equation for the line parallel to y =-2/3x – 7 that contains (-6, 3) is y - 3 = -2/3(x+6)
Learn more on equation of aline here: https://brainly.com/question/13763238
