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Maths Question
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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!!
A= 3b+2 over b+1 give formula for b Help!!
Answers
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For more on marking an answer as the "Best Answer", please visit our FAQ.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.
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.