Are the following pair of linear equations consistent? Justify your answer.
$2ax + by = a$
$4ax + 2by - 2a = 0$; $a, b \neq 0$

(A) pair of linear equations is consistent if it has at least one solution. The conditions are:
$1$. $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ (Unique solution)
$2$. $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ (Infinitely many solutions)
Given equations:
$2ax + by - a = 0$
$4ax + 2by - 2a = 0$
Here,$a_1 = 2a, b_1 = b, c_1 = -a$ and $a_2 = 4a, b_2 = 2b, c_2 = -2a$.
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{2a}{4a} = \frac{1}{2}$
$\frac{b_1}{b_2} = \frac{b}{2b} = \frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-a}{-2a} = \frac{1}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$,the system has infinitely many solutions. Therefore,the pair of linear equations is consistent.