Algebra for James

\[2a=b-c\left( \frac{b}{3} - 4 \right) \]

Multiply out the bracket: \[ 2a=b - \frac{c.b}{3} + 4c \]

Move terms containing \( b \) to one side: \[ \frac{c.b}{3} - b = 4c - 2a \]

Split \( b \) out as a factor on the LHS: \[ b \left( \frac{c}{3} - 1 \right) = 4c - 2a \]

Tidy up the LHS (i.e. express the bracketed term as a single fraction): \[ b \left( \frac{c - 3}{3} \right) = 4c - 2a \]

Multiply across the equation by \( \frac{3}{c-3} \) : \[ b = \frac{3}{c-3} \left( 4c - 2a \right) \]  

Finally leaving: \[ b =  6\left( \frac{2c - a}{c-3} \right) \]