Find the values of $ a $ and $ b $ for which the following system of linear equations has infinite number of solutions:
$2x-3y=7$
$(a+b)x-(a+b-3) y=4a+b$

Given: 

The given system of equations is:

$2x-3y=7$
$(a+b)x-(a+b-3) y=4a+b$

To do: 

We have to find the values of $a$ and $b$ for which the following system of linear equations has infinite number of solutions.

Solution:

The given system of equations can be written as:

$2x-3y-7=0$

$(a+b)x-(a+b-3) y-(4a+b)=0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

Comparing the given system of equations with the standard form of equations, we have,

$a_1=2, b_1=-3, c_1=-7$ and $a_2=(a+b), b_2=-(a+b-3), c_2=-(4a+b)$

The condition for which the given system of equations has infinitely many solutions is

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

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

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

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

$(a+b-3)\times2=3\times(a+b)$ and $(4a+b)\times2=7\times(a+b)$

$2a+2b-6=3a+3b$ and $8a+2b=7a+7b$

$3a-2a+3b-2b=-6$ and $8a-7a=7b-2b$

$a+b=-6$ and $a=5b$

Using $a=5b$ in $a+b=-6$, we get,

$5b+b=-6$

$6b=-6$

$b=-1$

This implies,

$a=5b=5(-1)=-5$

The values of $a$ and $b$ for which the given system of equations has infinitely many solutions is $-5$ and $-1$ respectively.