I am trying to follow the text in the appendix, however I get stuck when I come to the part where I need to solve for q1. As Far As I can see I need to find the inverse, which I have seen examples off when I know the function. But in this case I due not understand how this is solved when I do not know the function. Could someone please explain how this is solved for q1?
Before (6A.5), it says solving Eq.6A.4 for $q_1$. Carrying on from here: \begin{align*} & a-b(2q_1+q_2+\cdots+q_n) = m \\[0.1in] \implies & \frac{a-m}{b}=2q_1+q_2+\cdots+q_n \\[0.1in] \implies & \frac{a-m}{b}= 2q_1 +(n-1)q \quad \text{because } q_2=q_3=\cdots=q_n \\[0.1in] \implies & \frac{a-m}{b} -(n-1)q= 2q_1 \\ \implies & q_1=\frac{a-m}{2b} -\frac{n-1}{2}q. \end{align*}