1. **State the problem:** We are given the system of equations: $$5x + 14y = 45$$ $$10x + 7y = 27$$ We need to find the value of the product $xy$ where $(x,y)$ is the solution. 2. **Use the elimination method:** Multiply the first equation by 2 to align coefficients of $x$: $$2(5x + 14y) = 2(45) \Rightarrow 10x + 28y = 90$$ 3. **Subtract the second equation from this new equation:** $$\cancel{10x} + 28y - (\cancel{10x} + 7y) = 90 - 27$$ $$28y - 7y = 63$$ $$21y = 63$$ 4. **Solve for $y$:** $$y = \frac{63}{21} = 3$$ 5. **Substitute $y=3$ into the first original equation:** $$5x + 14(3) = 45$$ $$5x + 42 = 45$$ 6. **Solve for $x$:** $$5x = 45 - 42 = 3$$ $$x = \frac{3}{5}$$ 7. **Calculate the product $xy$:** $$xy = \frac{3}{5} \times 3 = \frac{9}{5}$$ **Final answer:** $$\boxed{\frac{9}{5}}$$