1. **State the problem:** Solve the system of equations: $$\begin{cases}-9a + b + 8c = -18 \\ -5a + 5b + 8c = -10 \\ -2a + 8b + 9c = 1 \end{cases}$$ 2. **Method:** We will use the elimination or substitution method to find $a$, $b$, and $c$. 3. **Step 1: Eliminate $c$ by subtracting equations.** Subtract equation 1 from equation 2: $$(-5a + 5b + 8c) - (-9a + b + 8c) = -10 - (-18)$$ Simplify: $$-5a + 5b + 8c + 9a - b - 8c = -10 + 18$$ $$4a + 4b = 8$$ Divide both sides by 4: $$\cancel{4}a + \cancel{4}b = \cancel{4}2$$ $$a + b = 2$$ 4. **Step 2: Eliminate $c$ by subtracting equations 3 and 1:** $$(-2a + 8b + 9c) - (-9a + b + 8c) = 1 - (-18)$$ Simplify: $$-2a + 8b + 9c + 9a - b - 8c = 1 + 18$$ $$7a + 7b + c = 19$$ Note: We made a mistake here; let's carefully subtract the $c$ terms: $$9c - 8c = c$$ So the equation is: $$7a + 7b + c = 19$$ 5. **Step 3: Express $b$ from step 3.** From step 3, we have $a + b = 2$, so: $$b = 2 - a$$ 6. **Step 4: Substitute $b$ into the third equation from step 4:** $$7a + 7(2 - a) + c = 19$$ Simplify: $$7a + 14 - 7a + c = 19$$ $$14 + c = 19$$ $$c = 5$$ 7. **Step 5: Substitute $c=5$ into the first equation:** $$-9a + b + 8(5) = -18$$ $$-9a + b + 40 = -18$$ $$-9a + b = -58$$ 8. **Step 6: Substitute $b = 2 - a$ into the above:** $$-9a + (2 - a) = -58$$ $$-9a + 2 - a = -58$$ $$-10a + 2 = -58$$ $$-10a = -60$$ $$a = 6$$ 9. **Step 7: Find $b$ using $b = 2 - a$:** $$b = 2 - 6 = -4$$ 10. **Final solution:** $$a = 6, b = -4, c = 5$$ This corresponds to option B.