1. **State the problem:** We have an initial alloy of silver and copper with unknown weight $x$ kg. 2. When $x$ kg of this alloy is mixed with 3 kg of pure silver, the new alloy contains 90% silver by weight. 3. Similarly, mixing $x$ kg of the initial alloy with 2 kg of another alloy that contains 90% silver results in an alloy with 84% silver. 4. **Define variables:** Let the fraction of silver in the initial alloy be $s$ (in decimal form). 5. **First mixture equation:** The total silver weight is from the initial alloy plus 3 kg pure silver, $$ \text{Silver}: xs + 3 $$ Total weight of the alloy is $$ x + 3 $$ So, $$ \frac{xs + 3}{x + 3} = 0.9 $$ 6. Multiply both sides by $x + 3$: $$ xs + 3 = 0.9x + 2.7 $$ Rearranged: $$ xs - 0.9x = 2.7 - 3 $$ $$ x(s - 0.9) = -0.3 $$ 7. **Second mixture equation:** Silver in $x$ kg of initial alloy plus silver in 2 kg of 90% alloy is $$ xs + 2 \times 0.9 = xs + 1.8 $$ Total weight: $$ x + 2 $$ Silver percent: $$ \frac{xs + 1.8}{x + 2} = 0.84 $$ 8. Multiply both sides by $x + 2$: $$ xs + 1.8 = 0.84x + 1.68 $$ Rearranged: $$ xs - 0.84x = 1.68 - 1.8 $$ $$ x(s - 0.84) = -0.12 $$ 9. **Solve the system:** From step 6, $$ x(s - 0.9) = -0.3 \Rightarrow s - 0.9 = \frac{-0.3}{x} $$ From step 8, $$ x(s - 0.84) = -0.12 \Rightarrow s - 0.84 = \frac{-0.12}{x} $$ 10. Subtract the two equations: $$ (s - 0.9) - (s - 0.84) = \frac{-0.3}{x} - \frac{-0.12}{x} $$ $$ -0.9 + 0.84 = \frac{-0.3 + 0.12}{x} $$ $$ -0.06 = \frac{-0.18}{x} $$ 11. Solve for $x$: $$ x = \frac{-0.18}{-0.06} = 3 $$ 12. **Therefore,** the weight of the initial alloy is $\boxed{3}$ kg. **Answer: Option C (3 kg).