1. **Problem Statement:** Find the individual concentrations of A, B, and C given the system: $$\begin{cases} a + b + c = 120 \\ 2a + 3b + c = 250 \\ a + 4b + 2c = 310 \end{cases}$$ 2. **Formula and Rules:** We solve this system of linear equations using substitution or elimination methods. 3. **Step 1: Write the system clearly:** $$a + b + c = 120 \quad (1)$$ $$2a + 3b + c = 250 \quad (2)$$ $$a + 4b + 2c = 310 \quad (3)$$ 4. **Step 2: Eliminate $c$ by subtracting (1) from (2) and (3):** From (2) - (1): $$ (2a + 3b + c) - (a + b + c) = 250 - 120 $$ $$ a + 2b = 130 \quad (4)$$ From (3) - (1): $$ (a + 4b + 2c) - (a + b + c) = 310 - 120 $$ $$ 3b + c = 190 \quad (5)$$ 5. **Step 3: Express $a$ from (4):** $$ a = 130 - 2b \quad (6)$$ 6. **Step 4: Substitute $a$ from (6) into (1):** $$ (130 - 2b) + b + c = 120 $$ $$ 130 - b + c = 120 $$ $$ c = 120 - 130 + b = b - 10 \quad (7)$$ 7. **Step 5: Substitute $c$ from (7) into (5):** $$ 3b + (b - 10) = 190 $$ $$ 4b - 10 = 190 $$ $$ 4b = 200 $$ $$ b = 50 $$ 8. **Step 6: Find $a$ and $c$ using $b=50$:** From (6): $$ a = 130 - 2(50) = 130 - 100 = 30 $$ From (7): $$ c = 50 - 10 = 40 $$ 9. **Final answer:** The individual concentrations are: $$ a = 30, \quad b = 50, \quad c = 40 $$