1. **Stating the problem:** We are given a system of three linear equations with three variables $A$, $B$, and $C$: $$ \begin{cases} A + B + C = 11000 \\ A + 2B + 3C = 25000 \\ 4A + 5B + 7C = 63000 \end{cases} $$ Our goal is to find the values of $A$, $B$, and $C$ that satisfy all three equations simultaneously. 2. **Method:** We will use the method of elimination or substitution to solve this system. 3. **Step 1: Express $A$ from the first equation:** $$ A = 11000 - B - C $$ 4. **Step 2: Substitute $A$ into the second and third equations:** Second equation: $$ (11000 - B - C) + 2B + 3C = 25000 $$ Simplify: $$ 11000 - B - C + 2B + 3C = 25000 $$ $$ 11000 + ( -B + 2B ) + ( -C + 3C ) = 25000 $$ $$ 11000 + B + 2C = 25000 $$ Subtract 11000 from both sides: $$ \cancel{11000} + B + 2C = 25000 - \cancel{11000} $$ $$ B + 2C = 14000 $$ 5. **Step 3: Substitute $A$ into the third equation:** $$ 4(11000 - B - C) + 5B + 7C = 63000 $$ Expand: $$ 44000 - 4B - 4C + 5B + 7C = 63000 $$ Combine like terms: $$ 44000 + ( -4B + 5B ) + ( -4C + 7C ) = 63000 $$ $$ 44000 + B + 3C = 63000 $$ Subtract 44000 from both sides: $$ \cancel{44000} + B + 3C = 63000 - \cancel{44000} $$ $$ B + 3C = 19000 $$ 6. **Step 4: Solve the system of two equations with two variables:** $$ \begin{cases} B + 2C = 14000 \\ B + 3C = 19000 \end{cases} $$ Subtract the first equation from the second: $$ (B + 3C) - (B + 2C) = 19000 - 14000 $$ $$ B - B + 3C - 2C = 5000 $$ $$ C = 5000 $$ 7. **Step 5: Substitute $C = 5000$ into $B + 2C = 14000$:** $$ B + 2(5000) = 14000 $$ $$ B + 10000 = 14000 $$ Subtract 10000 from both sides: $$ \cancel{10000} + B = 14000 - \cancel{10000} $$ $$ B = 4000 $$ 8. **Step 6: Substitute $B = 4000$ and $C = 5000$ into $A = 11000 - B - C$:** $$ A = 11000 - 4000 - 5000 = 11000 - 9000 = 2000 $$ **Final answer:** $$ A = 2000, \quad B = 4000, \quad C = 5000 $$