If the square of a-2b + (b-3c) is 0, what is the value of a + b-2c
∵ a-2b + (b-3c) square = 0,
And because a-2b ≥ 0 and (b-3c) 178; ≥ 0
∴a-2b=0,b-3c=0
A = 2B, B = 3C
∴a=6c
∴a+b-2c=6c+3c-2c=7c
∵ a-2b + (b-3c) square = 0,
And because a-2b ≥ 0 and (b-3c) 178; ≥ 0
∴a-2b=0,b-3c=0
A = 2B, B = 3C
∴a=6c
∴a+b-2c=6c+3c-2c=7c