1. **State the problem:** We are given the differential equation $$\frac{dy}{dx} = \frac{7 - 4x - 4y}{x + y - 3}$$ and we want to analyze or solve it. 2. **Identify the type of differential equation:** This is a first-order differential equation where the derivative $\frac{dy}{dx}$ is expressed as a ratio of linear expressions in $x$ and $y$. 3. **Rewrite the equation:** Let’s write it as $$\frac{dy}{dx} = \frac{7 - 4x - 4y}{x + y - 3}$$ 4. **Try substitution to simplify:** Define new variables to reduce the equation. Let $$u = x + y - 3$$ Then, $$y = u - x + 3$$ 5. **Find $\frac{dy}{dx}$ in terms of $u$ and $x$:** $$\frac{dy}{dx} = \frac{du}{dx} - 1$$ 6. **Rewrite numerator in terms of $u$ and $x$:** $$7 - 4x - 4y = 7 - 4x - 4(u - x + 3) = 7 - 4x - 4u + 4x - 12 = -4u - 5$$ 7. **Rewrite denominator in terms of $u$:** $$x + y - 3 = u$$ 8. **Substitute into the original equation:** $$\frac{dy}{dx} = \frac{-4u - 5}{u}$$ Using step 5, $$\frac{du}{dx} - 1 = \frac{-4u - 5}{u}$$ 9. **Solve for $\frac{du}{dx}$:** $$\frac{du}{dx} = 1 + \frac{-4u - 5}{u} = 1 - 4 - \frac{5}{u} = -3 - \frac{5}{u}$$ 10. **Rewrite as:** $$\frac{du}{dx} = -3 - \frac{5}{u}$$ 11. **Separate variables:** $$\frac{du}{dx} = -3 - \frac{5}{u} \implies \frac{du}{dx} = -\frac{3u + 5}{u}$$ 12. **Rewrite as:** $$\frac{du}{dx} = -\frac{3u + 5}{u}$$ 13. **Separate variables:** $$\frac{u}{3u + 5} du = -dx$$ 14. **Integrate both sides:** $$\int \frac{u}{3u + 5} du = - \int dx$$ 15. **Simplify the integral on the left:** Rewrite numerator: $$u = \frac{1}{3}(3u + 5) - \frac{5}{3}$$ So, $$\int \frac{u}{3u + 5} du = \int \frac{\frac{1}{3}(3u + 5) - \frac{5}{3}}{3u + 5} du = \int \left(\frac{1}{3} - \frac{5}{3(3u + 5)}\right) du$$ 16. **Integrate term by term:** $$= \int \frac{1}{3} du - \frac{5}{3} \int \frac{1}{3u + 5} du = \frac{u}{3} - \frac{5}{3} \cdot \frac{1}{3} \ln|3u + 5| + C = \frac{u}{3} - \frac{5}{9} \ln|3u + 5| + C$$ 17. **Integrate right side:** $$- \int dx = -x + C'$$ 18. **Combine constants and write implicit solution:** $$\frac{u}{3} - \frac{5}{9} \ln|3u + 5| = -x + C''$$ 19. **Recall substitution $u = x + y - 3$:** $$\frac{x + y - 3}{3} - \frac{5}{9} \ln|3(x + y - 3) + 5| = -x + C''$$ 20. **Final implicit solution:** $$\frac{x + y - 3}{3} - \frac{5}{9} \ln|3x + 3y - 9 + 5| = -x + C$$ This implicit equation relates $x$ and $y$ and represents the general solution to the differential equation.