Question

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

Answer 1

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SUM of number is
$(a + 3b - 4c) + (4a - b + 9c) + ( - 2b + 3c - a) \\ = a + 4a - a + 3b - b - 2b - 4c + 9c + 3c \\ = 4a + 8c$
Now Subtract
$(4a + 8c) - (2a - 3b + 4c) \\ = 4a - 2a + 3b + 8c - 4c \\ = 2a + 3b + 4c$
Hope it helps you

Answer 2

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