1. **State the problem:** Solve the system of linear equations: $$3x + 4y = 14$$ $$5x - 2y = 3$$ 2. **Choose a method:** We will use the elimination method to solve for $x$ and $y$. 3. **Eliminate one variable:** Multiply the first equation by 2 and the second equation by 4 to align coefficients of $y$: $$2(3x + 4y) = 2(14) \Rightarrow 6x + 8y = 28$$ $$4(5x - 2y) = 4(3) \Rightarrow 20x - 8y = 12$$ 4. **Add the two equations to eliminate $y$:** $$6x + 8y + 20x - 8y = 28 + 12$$ $$ (6x + 20x) + (8y - 8y) = 40$$ $$26x = 40$$ 5. **Solve for $x$:** $$x = \frac{40}{26} = \frac{20}{13}$$ 6. **Substitute $x$ back into one of the original equations to find $y$:** Using the first equation: $$3\left(\frac{20}{13}\right) + 4y = 14$$ $$\frac{60}{13} + 4y = 14$$ 7. **Isolate $y$:** $$4y = 14 - \frac{60}{13} = \frac{182}{13} - \frac{60}{13} = \frac{122}{13}$$ 8. **Solve for $y$:** $$y = \frac{122}{13} \times \frac{1}{4} = \frac{122}{52} = \frac{61}{26}$$ **Final answer:** $$x = \frac{20}{13}, \quad y = \frac{61}{26}$$