First we find the sum of (a + 3b – 4c), (4a – b + 9c) and (-2b + 3c – a)
= (a + 3b – 4c) + (4a – b + 9c) + (-2b + 3c – a)
= (a + 3b – 4c + 4a – b + 9c – 2b + 3c – a)
= (a + 4a – a) + (3b – b – 2b) + (-4c + 9c + 3c)
= (1 + 4 – 1)a + (3 – 1 – 2)b + (-4 + 9 + 3)c
= 4a + (0)b + 8c
=4a + 8c
Then,
Subtract (2a – 3b + 4c) from (4a + 8c)
= (4a + 8c) – (2a – 3b + 4c)
= 4a + 8c – 2a + 3b – 4c
= (4a – 2a) + (3b) + (8c – 4c)
= 2a + 3b + 4c
