Question

# Find the excess of $4(a - b - c)$ over $3(a + b)$ and subtract $9(a - 2b + bc)$ from the result. The final answer will bea. $( - 8a + 11b - 4c - 9bc)$b. $(8a + 11b - 4c - 9bc)$c. $( - 8a - 11b - 4c - 9bc)$d. $( - 8a - 11b + 4c - 9bc)$

Hint: In the first step, we subtract $3(a + b)$ from $4(a - b - c)$, and then subtract $9(a - 2b + bc)$ from the result we get, to get the final result.
Initially we are given the task to find the excess of $4(a - b - c)$ over $3(a + b)$. So, for that we will have to subtract $3(a + b)$ from $4(a - b - c)$,
$4(a - b - c) - 3(a + b) \\ = 4a - 4b - 4c - 3a - 3b \\ = a - 7b - 4c \\$
Now from the result we have to subtract $9(a - 2b + bc)$,
$(a - 7b - 4c) - 9(a - 2b + bc) \\ = a - 7b - 4c - 9a + 18b - 9bc \\ = - 8a + 11b - 4c - 9bc \\$
So, we have the answer as, $- 8a + 11b - 4c - 9bc$ which is option (a).