Subtaract (2a-3b+4c) from the sum of (a+3b-4c),(4a-b+9c)and (-2b+3c-a)
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So, First we will do the sum of the three given polynomials,
Sum =(a+3b−4c)+(4a−b+9c)+(−2b+3c−a)
=(a+4a−a)+(3b−b−2b)+(−4c+9c+3c)
=4a+8c
Now, we can perform the subtraction,
∴ Required difference
=(4a+8c)−(2a−3b+4c)
=4a+8c−2a+3b−4c
=2a+3b+4c
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