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secondary 4 | E Maths
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Can i present my answer as 2b-5A/A+b instead of 5A-2b/b+A
Date Posted: 3 years ago
Views: 191
cannot. its not the same
Nope it’s not the same. 5A has to be positive and 2b has to be negative
c = (5b - 2A)/(b + A)
Or
c = (5b - 2A)/(A + b)
For the denominator, a + b = b + a so the order does not matter.
BUT, for the numerator, 5b - 2A and 2A - 5b are the negatives of each other. They aren't the same.
Something similar would be like :
5 - 2 = 3 but 2 - 5 = -3, which ≠ 3
If you want to express the numerator as 2b - 5A, you have to change your denominator to the negative as well.
c = (5b - 2A)/(b + A)
= -(5b - 2A)/-(b + A)
= (-5b + 2A)/(-b - A)
= (2A - 5b)/(-b - A)
(You can skip the middle two steps, just showing for clarity.)
This is similar to :
c = 5/2 = -5/-2
But typically, we try to leave the expression with as few negative signs as possible, so the first two on top are preferred.
Or
c = (5b - 2A)/(A + b)
For the denominator, a + b = b + a so the order does not matter.
BUT, for the numerator, 5b - 2A and 2A - 5b are the negatives of each other. They aren't the same.
Something similar would be like :
5 - 2 = 3 but 2 - 5 = -3, which ≠ 3
If you want to express the numerator as 2b - 5A, you have to change your denominator to the negative as well.
c = (5b - 2A)/(b + A)
= -(5b - 2A)/-(b + A)
= (-5b + 2A)/(-b - A)
= (2A - 5b)/(-b - A)
(You can skip the middle two steps, just showing for clarity.)
This is similar to :
c = 5/2 = -5/-2
But typically, we try to leave the expression with as few negative signs as possible, so the first two on top are preferred.
See 1 Answer
Hi Meredy,
No. Because you are expressing c in terms of the other terms.
When you multiply the RHS with -ve,
you also need to multiply the LHS with -ve, which will give you: minus c
No. Because you are expressing c in terms of the other terms.
When you multiply the RHS with -ve,
you also need to multiply the LHS with -ve, which will give you: minus c
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