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Q.

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

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a

(−8a+11b−4c−9bc)

b

(8a+11b−4c−9bc)

c

(−8a−11b−4c−9bc)

d

(−8a−11b+4c−9bc)

answer is A.

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Detailed Solution

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,
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),
Then we have,
(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 (1).
.

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