1. **Stating the problem:** Solve the system of equations using elimination and substitution methods: $$3x + 2y = 12$$ $$x - y = 1$$ 2. **Using substitution method:** From the second equation, express $x$ in terms of $y$: $$x = y + 1$$ 3. Substitute $x = y + 1$ into the first equation: $$3(y + 1) + 2y = 12$$ 4. Simplify and solve for $y$: $$3y + 3 + 2y = 12$$ $$5y + 3 = 12$$ $$5y = 12 - 3$$ $$5y = 9$$ $$y = \frac{9}{5}$$ 5. Substitute $y = \frac{9}{5}$ back into $x = y + 1$: $$x = \frac{9}{5} + 1 = \frac{9}{5} + \frac{5}{5} = \frac{14}{5}$$ 6. **Using elimination method:** Multiply the second equation by 2 to align coefficients of $y$: $$2(x - y) = 2(1)$$ $$2x - 2y = 2$$ 7. Write the system: $$3x + 2y = 12$$ $$2x - 2y = 2$$ 8. Add the two equations to eliminate $y$: $$3x + 2y + 2x - 2y = 12 + 2$$ $$5x = 14$$ $$x = \frac{14}{5}$$ 9. Substitute $x = \frac{14}{5}$ into the second original equation: $$\frac{14}{5} - y = 1$$ 10. Solve for $y$: $$-y = 1 - \frac{14}{5} = \frac{5}{5} - \frac{14}{5} = -\frac{9}{5}$$ $$y = \frac{9}{5}$$ **Final answer:** $$x = \frac{14}{5}, \quad y = \frac{9}{5}$$