If $2x-3y=7$ and $(a+b)x-(a+b-3)y=4a+b$ represent coincident lines then a and b satisfy the equation

The correct option is C: $a-5b=0$
$2x-3y=7$ and $(a+b)x-(a+b-3)y=4a+b$ represent coincident lines then,

Condition for coincident lines, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, we can also write like this:

$\frac{a_2}{a_1}=\frac{b_2}{b_1}=\frac{c_2}{c_1}$

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

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

$\Rightarrow \frac{a+b}{2}=\frac{4a+b}{7}$

$\Rightarrow 7(a+b)=2(4a+b)$

$\Rightarrow 7a+7b=8a+2b$

$\therefore a-5b=0$

So, the correct option is C: $a-5b=0$