1. **State the problem:** Solve the system of linear equations: $$\frac{1}{2}x + \frac{2}{3}y = 1$$ $$\frac{1}{4}x - \frac{1}{3}y = \frac{5}{2}$$ 2. **Use the elimination or substitution method. Here, we use elimination.** 3. Multiply the first equation by 12 (the least common multiple of denominators 2 and 3) to clear fractions: $$12 \times \left(\frac{1}{2}x + \frac{2}{3}y\right) = 12 \times 1$$ $$6x + 8y = 12$$ 4. Multiply the second equation by 12 (LCM of 4 and 3) to clear fractions: $$12 \times \left(\frac{1}{4}x - \frac{1}{3}y\right) = 12 \times \frac{5}{2}$$ $$3x - 4y = 30$$ 5. Now the system is: $$6x + 8y = 12$$ $$3x - 4y = 30$$ 6. Multiply the second equation by 2 to align coefficients of $x$: $$2 \times (3x - 4y) = 2 \times 30$$ $$6x - 8y = 60$$ 7. Subtract the first equation from this new equation: $$\cancel{6x} - 8y - (\cancel{6x} + 8y) = 60 - 12$$ $$-16y = 48$$ 8. Solve for $y$: $$y = \frac{48}{-16} = -3$$ 9. Substitute $y = -3$ into the first simplified equation: $$6x + 8(-3) = 12$$ $$6x - 24 = 12$$ 10. Solve for $x$: $$6x = 12 + 24 = 36$$ $$x = \frac{36}{6} = 6$$ **Final answer:** $$x = 6, \quad y = -3$$