Calculus with Applications (10th Edition)

$$\frac{{dy}}{{dx}} = - 20 - \frac{{50}}{3}x$$
\eqalign{ & \frac{{6 + 5x}}{{2 - 3y}} = \frac{1}{{5x}} \cr & {\text{cross product}} \cr & 5x\left( {6 + 5x} \right) = 2 - 3y \cr & 60x + 25{x^2} = 2 - 3y \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {60x + 25{x^2}} \right) = \frac{d}{{dx}}\left( {2 - 3y} \right) \cr & {\text{solve the derivatives }} \cr & 60 + 50x = 0 - 3\frac{{dy}}{{dx}} \cr & {\text{solve for }}\frac{{dy}}{{dx}} \cr & 60 + 50x = 0 - 3\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = \frac{{60 + 50x}}{{ - 3}} \cr & \frac{{dy}}{{dx}} = - 20 - \frac{{50}}{3}x \cr}