Question

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

$2x-3y=7$

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

Answer 1

Given $2x-3y=7$

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

To have infinite no. of solutions ,

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

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

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

$a=5b-----\left(1\right)$

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

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

$5a-4b+21=0-------\left(2\right)$

From (1) & (2)

$25b-4b+21=0$

$b=-1$

$a=-5$