1. **Stating the problem:** Solve the system of linear equations: $$\frac{1}{2}x + 3y = 3$$ $$\frac{1}{6} + 4y = 3$$ 2. **Analyze the second equation first:** $$\frac{1}{6} + 4y = 3$$ Subtract $\frac{1}{6}$ from both sides: $$4y = 3 - \frac{1}{6}$$ 3. **Simplify the right side:** $$3 = \frac{18}{6}$$ So, $$4y = \frac{18}{6} - \frac{1}{6} = \frac{17}{6}$$ 4. **Solve for $y$ by dividing both sides by 4:** $$y = \frac{\frac{17}{6}}{4} = \frac{17}{6} \times \frac{1}{4} = \frac{17}{24}$$ 5. **Substitute $y = \frac{17}{24}$ into the first equation:** $$\frac{1}{2}x + 3 \times \frac{17}{24} = 3$$ Simplify: $$\frac{1}{2}x + \frac{51}{24} = 3$$ 6. **Subtract $\frac{51}{24}$ from both sides:** $$\frac{1}{2}x = 3 - \frac{51}{24}$$ Convert 3 to $\frac{72}{24}$: $$\frac{1}{2}x = \frac{72}{24} - \frac{51}{24} = \frac{21}{24}$$ 7. **Solve for $x$ by dividing both sides by $\frac{1}{2}$:** $$x = \frac{\frac{21}{24}}{\frac{1}{2}} = \frac{21}{24} \times 2 = \frac{42}{24}$$ Simplify the fraction: $$x = \frac{42}{24} = \frac{7}{4}$$ **Final answer:** $$x = \frac{7}{4}, \quad y = \frac{17}{24}$$