## College Algebra (10th Edition)

$\color{blue}{y=-\dfrac{2}{3}x-\dfrac{1}{3}}$
RECALL: The slope-intercept form of a line's equation is: $y=mx+b$ where $m$=slope and $b$ = y-intercept. The line has a slope of $-\dfrac{2}{3}$, so 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 given in the problem into the tentative equation above to obtain: $y=-\dfrac{2}{3}x+b \\-1=-\dfrac{2}{3}(1)+b \\-1=-\dfrac{2}{3}+b \\-1+\dfrac{2}{3}=b \\-\dfrac{3}{3}+\dfrac{2}{3}=b \\-\dfrac{1}{3}=b$ Therefore, the equation of the line is $\color{blue}{y=-\dfrac{2}{3}x-\dfrac{1}{3}}$.