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Maths Question

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quizzkid | 14:31 Sun 01st Feb 2015 | Quizzes & Puzzles
6 Answers
Hi I need help with this maths question please. I never was good at algebra.
A= 3b+2 over b+1 give formula for b Help!!
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A = 3b + 2/ b+1
multiply both sides by b+1
gives Ab + A = 3b +2
rearranging
gives Ab - 3b = 2 - A
factorise RHS b(A-3) = 2 - A
now divide both sides by A-3
gives b = (2-A) over (A-3)
First thing to do whenever you are presented with a fraction is to get rid of the fraction. Here the way to do that is multiply both sides by (b+1), giving:

A*(b+1) = [(3b+2)/(b+1)]*(b+1)

On the right-hand side the b+1 on the top and bottom of the fraction cancel, leaving just 3b+2. So:

A*(b+1) = (3b+2)

Now expand the brackets on the left, as A*(b+1) = A*b + A*1 = A*b + A, leaving:

A*b + A = 3b+2

Next, we want to rearrange for b, so move everything that has a b in it to one side, and anything that doesn't have a b in it to the other. Here we should subtract A from both sides, and also subtract 3b from both sides, cancelling the A on the left and the 3b on the right:

A*b + A - A - 3b = 3b - 3b + 2 - A

or A*b - 3b = 2-A

Now just as A(b+1) = A*b + A, we can replace A*b - 3b by b*(A-3) (you can see this be expanding the brackets again). Hence the equation becomes

b*(A-3) = 2-A

Finally, we must divide both sides by A-3, removing it from the left-hand side, and leaving:

b*(A-3)/(A-3) = (2-A)/(A-3)

or just

b = (2-A)/(A-3)

Guiding principle: do whatever you want to both sides of the equation, but make sure you do exactly the same thing on both sides. Hence add A to both, or subtract A from both, or divide both sides by b or multiply both sides by 2, but always do the same thing on both sides.
(b+1)A=3b+2
bA +A= 3b+2
bA-3b= 2-A
b(A-3)= 2-A

b= (2-A)/(a-3)

Same as Fibaonacci
Sorry jim, I hadn't seen yours
Question Author
Thank you all so much for your help. Hopefully I will be able to sort out the rest of them from your explanations.
You're welcome and do feel free to ask any other questions you might have. Hope the explanations are useful, though.

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