Simple Algebra Equation I have a simple part of a question to solve.  The problem is my answer is different to the solution in my textbook.
The equation is: $$\frac{5v}{6} = \frac{(\frac{1}{2}a+b+\frac{1}{2} c)v}{a+b+c}$$
I am supposed to get $$\frac{2}{3}(a+b+c) = b$$
But I simply get: $$b=2a +2c$$
I get my answer by cross multiplying.  I then use my answer to get $\frac{b}{a+b+c}$ as some fraction.  I have not worked onto this stage as I am unsure about the above work.
What am I doing wrong here? 

$b=2a+2c$
Adding $2b$ on both sides gives
$b + 2b =2a+2b+2c$
Or better
$3b =2a+2b+2c$
And if you throw the $3$ to the other side you get:
$b =\frac{2}{3}(a+b+c)$
