Rearranging formulae

[Skelly4444]

I am having trouble rearranging the following formula in order to get (b) as the subject.

(2a/3b) - b = 4c + 1

Any guidance would be most appreciated

 

[Deleted member 4993]

I am having trouble rearranging the following formula in order to get (b) as the subject.

(2a/3b) - b = 4c + 1

Any guidance would be most appreciated

This will become a quadratic equation. What have you been taught about solution of quadratic equation?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this problem.

 

[Skelly4444]

I seem to end up with a quadratic of the following form.

3b + 4bc + b = 2a/3

I cannot reduce the quadratic by completing the square or factorising so now I am wondering if the question is possible?

 

[Skelly4444]

There was meant to be a squared sign on the 3b but it hasn't displayed

 

[Skelly4444]

Sorry, it meant to read 3b^2 + 12bc + 3b = 2a/3

 

[HallsofIvy]

First, 4bc+ b= (4c+ 1)b. The next thing I would do is subtract 2a/3 from both sides to get 3b^2+ (4c+ 1)b- 2a/3= 0. Now, do you know and can you use the "quadratic formula"?

 

[Skelly4444]

That's great, thanks for your help, I can do it from here by completing the square.

 

[lookagain]

I am having trouble rearranging the following formula in order to get (b) as the subject.

(2a/3b) - b = 4c + 1

Any guidance would be most appreciated

This is written incorrectly if b is meant to be in the denominator.
The parentheses need to go around the denominator because
of the Order of Operations:

2a/(3b) - b = 4c + 1

Then

(3b)[2a/(3b) - b] = (3b)[4c + 1]

\(\displaystyle 2a - 3b^2 \ = \ 12bc + 3b\)

\(\displaystyle 0 \ = \ 3b^2 + 12bc + 3b - 2a\)

or

\(\displaystyle 3b^2 + (12c + 3)b - 2a \ = \ 0 \)

Then you could use the Quadratic Formula without
substituting a fractional expression.


I would like to see how the problem looked, not necessarily how
you presented it to us (how you typed it out). It's the placement
of the b variable that concerns me, and it changes the difficulty
level/intent of the problem.