Find the values of a and b for which the system of linear equations has an infinite number of solutions: $(a-1) x+3 y=2$, $6 x+(1-2 b) y=6$

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Find the values of a and b for which the system of linear equations has an infinite number of solutions:
$(a-1) x+3 y=2$ ,
$6 x+(1-2 b) y=6$

Find the values of a and b for which the system of linear equations has an infinite number of solutions:
$(a-1) x+3 y=2$ ,
$6 x+(1-2 b) y=6$

Solution:

We can write the given system of equations as
$\begin{array}{l} (a-1) x+3 y=2 \\ \Rightarrow(a-1) x+3 y-2=0\dots \dots(i) \\ \text { and } 6 x+(1-2 b) y=6 \end{array}$
$\Rightarrow 6 \mathrm{x}+(1-2 \mathrm{~b}) \mathrm{y}-6=0\dots \dots(ii)$
The given equations are of the following form:
$\begin{array}{l} a_{1} x+b_{1} y+c_{1}=0 \\ a_{2} x+b_{2} y+c_{2}=0 \end{array}$
where, $a_{1}=(a-1), b_{1}=3, c_{1}=-2$ and $a_{2}=6, b_{2}=(1-2 b), c_{2}=-6$
For an infinite no. of solutions, we must have:
$\begin{array}{l} \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \\ \Rightarrow \frac{a-1}{6}=\frac{3}{(1-2 b)}=\frac{-2}{-6} \\ \Rightarrow \frac{a-1}{6}=\frac{3}{(1-2 b)}=\frac{1}{3} \\ \Rightarrow \frac{a-1}{6}=\frac{1}{3} \text { and } \frac{3}{(1-2 b)}=\frac{1}{3} \\ \Rightarrow 3 \mathrm{a}-3=6 \text { and } 9=1-2 \mathrm{~b} \\ \Rightarrow 3 \mathrm{a}=9 \text { and } 2 \mathrm{~b}=-8 \\ \Rightarrow \mathrm{a}=3 \text { and } \mathrm{b}=-4 \\ \quad \therefore \mathrm{a}=3 \text { and } \mathrm{b}=-4 \end{array}$