1. **State the problem:** Solve the system of linear equations: $$\begin{cases} 0.05x + 0.04y + 0.045z = 0 \\ 0.05x + 0.04y + 1.045z + w = 0 \\ 1.05x + 1.04y = 0 \end{cases}$$ 2. **Analyze the system:** There are 4 variables ($x, y, z, w$) but only 3 equations, so we expect infinitely many solutions depending on one free variable. 3. **From the third equation:** $$1.05x + 1.04y = 0$$ Solve for $x$: $$x = -\frac{1.04}{1.05} y$$ 4. **Substitute $x$ into the first equation:** $$0.05\left(-\frac{1.04}{1.05} y\right) + 0.04y + 0.045z = 0$$ Simplify: $$-0.05 \times \frac{1.04}{1.05} y + 0.04y + 0.045z = 0$$ Calculate coefficient: $$-\frac{0.05 \times 1.04}{1.05} y + 0.04y + 0.045z = 0$$ $$-\frac{0.052}{1.05} y + 0.04y + 0.045z = 0$$ $$-0.0495238 y + 0.04y + 0.045z = 0$$ Combine $y$ terms: $$(-0.0495238 + 0.04) y + 0.045z = 0$$ $$-0.0095238 y + 0.045z = 0$$ 5. **Solve for $z$ in terms of $y$:** $$0.045z = 0.0095238 y$$ $$z = \frac{0.0095238}{0.045} y = 0.21164 y$$ 6. **Substitute $x$ and $z$ into the second equation:** $$0.05x + 0.04y + 1.045z + w = 0$$ Replace $x$ and $z$: $$0.05 \left(-\frac{1.04}{1.05} y\right) + 0.04y + 1.045 \times 0.21164 y + w = 0$$ Calculate each term: $$-0.0495238 y + 0.04y + 0.2211 y + w = 0$$ Combine $y$ terms: $$(-0.0495238 + 0.04 + 0.2211) y + w = 0$$ $$0.211576 y + w = 0$$ 7. **Solve for $w$:** $$w = -0.211576 y$$ 8. **Summary of solutions:** $$x = -\frac{1.04}{1.05} y \approx -0.9905 y$$ $$z = 0.21164 y$$ $$w = -0.211576 y$$ $$y = y \quad \text{(free parameter)}$$ **Final answer:** The solution set is $$\boxed{\left(x, y, z, w\right) = \left(-0.9905 y, y, 0.21164 y, -0.211576 y\right) \text{ for any real } y}$$