Answer
$\color{blue}{y=\dfrac{2}{3}x+\dfrac{19}{3}}$
Work Step by Step
RECALL:
(1) The slope-intercept form of a line's equation is:
$y=mx+b$
where $m$ = slope and $b$ = y-intercept
(2) Parallel lines have equal slopes.
The line we are looking for is parallel to the line $\\2x-3y=-4$. Converting this equation to slope-intercept form gives:
$2x-3y=-4
\\-3y=-2x-4
\\\dfrac{-3y}{-3} = \dfrac{-2x-4}{-3}
\\y=\dfrac{2}{3}x+\dfrac{4}{3}$
The slope of this line is $\dfrac{2}{3}$.
This means that the slope of the line parallel to this line is also $\dfrac{2}{3}$.
Thus, the tentative equation of the line is:
$y=\dfrac{2}{3}x + b$
To find the value of $b$, substitute the x and y values of the point $(-5, 3)$ into the tentative equation above to obtain:
$y=\dfrac{2}{3}x+b
\\3=\dfrac{2}{3} \cdot (-5) + b
\\3 = -\dfrac{10}{3} +b
\\3 + \dfrac{10}{3} = b
\\\dfrac{9}{3} + \dfrac{10}{3} = b
\\\dfrac{19}{3} = b$
Thus, the equation of the line is $\color{blue}{y=\dfrac{2}{3}x+\dfrac{19}{3}}$.
