1. **State the problem:** We are given a system of two linear equations: $$x + y = 15$$ $$0.40x + 0.20y = 15$$ We need to find the values of $x$ and $y$ that satisfy both equations. 2. **Explain the formulas and rules:** This is a system of linear equations. We can solve it using substitution or elimination. Here, elimination is straightforward. 3. **Solve the system:** From the first equation: $$y = 15 - x$$ Substitute into the second equation: $$0.40x + 0.20(15 - x) = 15$$ 4. **Simplify the second equation:** $$0.40x + 3 - 0.20x = 15$$ Combine like terms: $$0.40x - 0.20x + 3 = 15$$ $$0.20x + 3 = 15$$ 5. **Isolate $x$:** $$0.20x = 15 - 3$$ $$0.20x = 12$$ 6. **Solve for $x$:** $$x = \frac{12}{0.20}$$ Show cancellation: $$x = \frac{12}{\cancel{0.20}} \times \frac{\cancel{1}}{1} = 60$$ 7. **Find $y$ using $y = 15 - x$:** $$y = 15 - 60 = -45$$ 8. **Interpretation:** The solution is $x = 60$ and $y = -45$. This means the values satisfy both equations, but $y$ is negative, which may or may not be meaningful depending on the context. **Final answer:** $$x = 60, \quad y = -45$$