subtract [c-3d-e]from the sum of [4c+d-e] and [2c-3d+2e]

Answer 1

To subtract [c-3d-e] from the sum of [4c+d-e] and [2c-3d+2e], first simplify the sum:

[4c+d-e] + [2c-3d+2e] = 6c - 2d + e

Then, subtract [c-3d-e]:

(6c - 2d + e) - (c - 3d - e) = 6c - 2d + e - c + 3d + e

Simplifying further:

6c - c - 2d + 3d + e + e = 5c + d + 2e

Therefore, the answer is 5c + d + 2e.