Subtract (2a − 3b + 4c) from the sum of (a + 3b − 4c), (4a − b + 9c) and (−2b + 3c − a).

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Solution

(a + 3b − 4c) + (4a − b + 9c) + (−2b + 3c − a)
= a + 4a - a + 3b -b -2b -4c +9c + 3c
= 4a + 8c

Now, (4a + 8c ) - (2a − 3b + 4c)
= 4a - 2a + 3b + 8c - 4c
= 2a + 3b + 4c

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