1. **State the problem:** Solve the system of equations: $$100 \times 10^{-6} = \frac{x}{180 \times 10^{6}} - \frac{0.28}{180 \times 10^{6}} y$$ $$-200 \times 10^{-6} = \frac{y}{180 \times 10^{6}} - \frac{0.28}{180 \times 10^{6}} x$$ 2. **Rewrite the equations for clarity:** Multiply both sides by $180 \times 10^{6}$ to clear denominators: $$100 \times 10^{-6} \times 180 \times 10^{6} = x - 0.28 y$$ $$-200 \times 10^{-6} \times 180 \times 10^{6} = y - 0.28 x$$ 3. **Calculate the left sides:** $$100 \times 10^{-6} \times 180 \times 10^{6} = 100 \times 180 = 18000$$ $$-200 \times 10^{-6} \times 180 \times 10^{6} = -200 \times 180 = -36000$$ 4. **Rewrite the system:** $$18000 = x - 0.28 y$$ $$-36000 = y - 0.28 x$$ 5. **Express as linear system:** $$x - 0.28 y = 18000$$ $$-0.28 x + y = -36000$$ 6. **Use substitution or elimination:** From first equation: $$x = 18000 + 0.28 y$$ Substitute into second: $$-0.28 (18000 + 0.28 y) + y = -36000$$ 7. **Simplify:** $$-0.28 \times 18000 - 0.28 \times 0.28 y + y = -36000$$ $$-5040 - 0.0784 y + y = -36000$$ $$-5040 + 0.9216 y = -36000$$ 8. **Solve for $y$:** $$0.9216 y = -36000 + 5040 = -30960$$ $$y = \frac{-30960}{0.9216} \approx -33600$$ 9. **Find $x$:** $$x = 18000 + 0.28 \times (-33600) = 18000 - 9408 = 8592$$ **Final answer:** $$x \approx 8592, \quad y \approx -33600$$