Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$y=\frac{5}{3}x+5$
RECALL: The slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the line's y-intercept. The line in Exercise 16 has $(0, 5)$ as its y-intercept so $b=5$. Thus, the tentative equation of the line is $y=mx + 5$. The line passes through the point $(-3, 0)$. To find the value of $m$, substitute the x and y coordinates of this point into the tentative equation above to obtain: $y=mx+5 \\0=m(-3)+5 \\0=-3m+5 \\0-5=-3m \\-5 = -3m \\\frac{-5}{-3}=\frac{-3m}{-3} \\\frac{5}{3} = m$ Therefore, the equation of the line in Exercise 16 is: $y=\frac{5}{3}x+5$