Maths Equation?

Question

I was given a question like this on a mathematics revision sheet.
Solve this equation for 'm'

A= b/4[4c+(b-2)m]

I have forgotten how to do this over the holidays.
Could you please give me detailed WORKING OUT with the answer as I would like to remember how to do these types of questions.

Answer 1

A = (b/4) [4c - (b - 2)m]

Multiply both sides by [4c - (b - 2)m],

A[4c - (b - 2)m] = b/4

Multiply both sides by 1/A. This will cancel out the A on the left hand side.

4c - (b - 2)m = (b/4)(1/A)

Merge the fractions on the right hand side by multiplying them.

4c - (b - 2)m = b/[4A]

Move the 4c to the right hand side.

-(b - 2)m = (b/[4A]) - 4c

Multiply both sides by (-1).

(b - 2)m = 4c - (b/[4A])

Now, divide both sides by (b - 2)

m = [4c - (b/[4A])] / (b - 2)

Answer 2

Have to get m all by itself. Divide both sides by (b/4) first, giving:
a/(b/4)=[4c+(b-2)m]
Clean it up a little, giving:
4a/b=4c+(b-2)m {just fixed the left side and got rid of the braces}
Now move over the 4c to get:
(4a/b)-4c=(b-2)m {added the parens on the left for clarity}
then move the (b-2) term, to get:
[(4a/b)-4c]/(b-2)=m {done. victory dance now}

The general technique is to remember that as long as you do the same thing to both sides of the = sign, all is well. so we just shuffle pieces around until we get the one we want all by itself.

Answer 3

That looks a bit like the sum of a geometric series.
Sn = n/2(2a-(n-1)d)
I'm not sure if that's it exactly, but a is the first term, d is the difference and n's are, well, the terms.
:)
Hope that helps.
If not, basically, just keep adding/subtracting/multiplying/dividing terms off the right handside onto the left. For instance, you could start with x by 4
Then you'd get
4A=b[4c+(b-2)m]
Then divide by b
(4A)/b=[4c+(b-2)m]
etc. :)
Luce xxxxx

Answer 4

[(4A/b)-4c-bm]/-2