Question

if a=(3a2-4b-1) B= (6a2+3b-8)
c=(4a2-9b+3) then find the value of A-B+C

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Answer 1

Answer:

$A = 3 {a}^{2} - 4b - 1 \\ B = 6 {a}^{2} + 3b - 8 \\ C = 4 {a}^{2} - 9b + 3 \\ \\ A - B + C \\ (3 {a}^{2} - 4b - 1) - (6 {a}^{2} + 3b - 8) + (4 {a}^{2} - 9b + 3) \\ = 3 {a}^{2} - 4b - 1 - 6 {a}^{2} - 3b + 8 + 4 {a}^{2} - 9b + 3 \\ = 3 {a}^{2} - 6 {a}^{2} + 4 {a}^{2} - 4b - 3b - 9b - 1 + 8 + 3 \\ = \underline{ \underline{ {a}^{2} - 16b + 10}}$