Find the values of a and b for which the following system of linear equations has infinite number of solutions :

GIVEN: 

$2 x-3 y=7$

$(a+b) x-(a+b-3) y=4 a+b$

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

$\frac{2}{(a+b)}=\frac{3}{(a+b-3)}=\frac{7}{4 a+b}$

Consider the following

$\frac{3}{(a+b-3)}=\frac{7}{4 a+b}$

$12 a+3 b=7(a+b-3)$

$12 a+3 b=7 a+7 b-21$

$5 a-4 b+21=0 \ldots \ldots(1)$

Again

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

$2(a+b-3)=3(a+b)$

$2 a+2 b-6=3 a+3 b$

$a+b+6=0 \ldots \ldots(2)$

Multiplying eq. (2) by 4 and adding eq.(1)

$9 a+45=0$

$a=-5$

Putting the value of a in eq.(2)

$-5+b+6=0$

$b=-1$

Hence for $a=-5$ and $b=-1$ the system of equation has infinitely many solution.