Determine the values of a and b so that the following system of linear equations have infinitely many solutions :

GIVEN: 

$(2 a-1) x+3 y-5=0$

$3 x+(b-1) y-2=0$

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution 

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Here

$\frac{(2 a-1)}{3}=\frac{3}{(b-1)}=\frac{5}{2}$

$\frac{3}{(b-1)}=\frac{5}{2}$

$6=5(b-1)$

$6=5 b-5$

$b=\frac{11}{5}$

Again consider

$\frac{(2 a-1)}{3}=\frac{3}{(b-1)}$

$(2 a-1)(b-1)=9$

$(2 a-1)\left(\frac{11}{5}-1\right)=9$ [subsituting the value of $b$ ]

$(2 a-1)\left(\frac{11-5}{5}\right)=9$

$(2 a-1)\left(\frac{6}{5}\right)=9$

$(2 a-1)=9\left(\frac{5}{6}\right)$

$(2 a-1)=\left(\frac{15}{2}\right)$

$2 a=\frac{15}{2}+1$

$2 a=\frac{15+2}{2}$

$2 a=\frac{17}{2}$

$a=\frac{17}{4}$

Hence for $a=\frac{17}{4}$ and $b=\frac{11}{5}$ the system of equation has infinitely many solution.