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 be
a. $$(-8a+11b-4c-9bc)$$
b. $$(8a+11b-4c-9bc)$$
c. $$(-8a-11b-4c-9bc)$$
d. $$(-8a-11b+4c-9bc)$$

Answer 1

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.

Complete step by step answer:

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 $$4(a-b-c)$$,
So by subtracting we get,
$$\begin{gathered} 4(a-b-c)-3(a+b) \\ =4a-4b-4c-3a-3b \\ =a-7b-4c \end{gathered}$$
Now from the result we have to subtract $$9(a-2b+bc)$$,
Then we have,
$$\begin{gathered} (a-7b-4c)-9(a-2b+bc) \\ =a-7b-4c-9a+18b-9bc \\ =-8a+11b-4c-9bc \end{gathered}$$
So, we have the answer as, $$-8a+11b-4c-9bc$$ which is option (a).

Note: If we talk about the excess term here, it is a bit confusing to deal with. The excess over a term means how much bigger the first term is than the 2nd term here. So, we need to take care of that a bit to get the correct answer.