# [SOLVED] Rearrange to solve

• May 11th 2009, 03:19 PM
marclurr
[SOLVED] Rearrange to solve
Okay I feel dumb asking this, I should know this, but it's been so long since I've bothered to use any algebra I've completely forgotten how it would be done.

Consider this equation.

a = b / (b + c)

Pretend we have a value for a and c, how would you rearrange this to find b? It being on both the top and bottom of the division is what's confusing me.

I did this but I know it's completely wrong:

a * c = b /b

but that would mean

a * c = 1

!!!

Can somebody point out what I'm doing wrong?

Cheers
• May 11th 2009, 03:22 PM
e^(i*pi)
Quote:

Originally Posted by marclurr
Okay I feel dumb asking this, I should know this, but it's been so long since I've bothered to use any algebra I've completely forgotten how it would be done.

Consider this equation.

a = b / (b + c)

Pretend we have a value for a and c, how would you rearrange this to find b? It being on both the top and bottom of the division is what's confusing me.

I did this but I know it's completely wrong:

a * c = b /b

but that would mean

a * c = 1

!!!

Can somebody point out what I'm doing wrong?

Cheers

Those brackets mean you have to multiply a by both terms inside it:

a*(b+c) = ab+ac
• May 11th 2009, 03:26 PM
marclurr
Would it make any difference if the brackets werent there? I just included them to make it clear that they were both on the 'bottom' of the division

cheers
• May 11th 2009, 03:27 PM
pickslides
Hi there, consider these steps.

$a=\frac{b}{b+c}$

I like to flip both sides to get

$\frac{1}{a}=\frac{b+c}{b}$

now working on the RHS

$\frac{1}{a}=\frac{b}{b}+\frac{c}{b}$

$\frac{1}{a}=1+\frac{c}{b}$

now we only have 1 b so we can rearrange, have a go from here....

$b=\frac{c}{\frac{1}{a}-1}$