## College Algebra (10th Edition)

$\color{blue}{y=\dfrac{2}{3}x+\dfrac{19}{3}}$
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 $\color{blue}{y=\dfrac{2}{3}x+\dfrac{19}{3}}$.