## College Algebra (6th Edition)

y=-$\frac{2}{3}$x-2
3x-2y-4=0 2y=3x-4 y=$\frac{3}{2}$x-2 Therefore the desired line has a slope of -$\frac{2}{3}$ and contains the point (0, -2). y-y$_1$=m(x-x$_1$) y+2=-$\frac{2}{3}$(x-0) y=-$\frac{2}{3}$x-2