Following a basic example from Wikipedia, I'd appreciate some help in understanding how the formulation was linearized. Would appreciate if someone could write it out explicitly. I understand the denominator of the objective function (OF) was set to 1. (If I understand correctly, this is the only substitution required to make all transformations).
$$10v_1+2v_2 = 1$$
So, using this substitution, the OF can be converted to a linear form
$$ \begin{align} \text{Non-linear} && \text{Linear} \\ \\ \frac{100u_1}{10v_1+2v_2} && 100u_1 \end{align} $$
And the same with the corresponding efficiency constraint for this unit:
$$ \begin{align} \text{Non-linear} && && \text{Linear} \\ \\ \frac{100u_1}{10v_1+2v_2}\leqslant 1 && && 100u_1 - (10v_1+2v_2)\leqslant 0 \end{align} $$
But how are the efficiency constraints of other units converted when their denominators are not equal to 1?
$$ \begin{align} \text{Non-linear} && && \text{Linear} \\ \\ \frac{80u_1}{8v_1+4v_2}\leqslant 1 && && 80u_1 - (8v_1+4v_2)\leqslant 0 \end{align} $$
Consider the non-linear inequality: $$\frac{80u_1}{8v_1+4v_2}\leq 1 \Rightarrow 80u_1\leq 8v_1+4v_2 \Rightarrow 80u_1-(8v_1+4v_2)\leq 0.$$ The above is straight forward. Now consider the linear inequality on the RHS $$80u_1-(8v_1+4v_2)\leq 0 \Rightarrow 80u_1\leq 8v_1+4v_2 \Rightarrow \frac{80u_1}{8v_1+4v_2}\leq 1$$ This implication holds true under the assumption that the denominator term $8v_1+4v_2 \neq 0.$ We can show that it is indeed true, by solving the following two linear inequalities: $$8v_1+4v_2 = 0 \text{ and } 10v_1+2v_2=1.$$ On solving, we get $v_2<0$ (Not possible). Hence, our assumption holds true. Similarly, we can show for other inequality also. Hope this helps.