Question

If $2x-3y=7$ and $(a+b)x-(a+b-3)y=4a+b$ have infinite solutions then $(a,b)=$

• A. $(-5,-1)$
• B. $(-5,1)$
• C. $(5,1)$
• D. $(5,-1)$

Answer 1

The correct option is A $(-5,-1)$

The equations are
$2x-3y-7=0$
$(a+b)x-(a+b-3)y-(4a+b)=0$
Here, $a_1=2,b_1=-3,c_1=-7$
$a_2=a+b,b_2=-(a+b-3),c_2=-(4a+b)$
The system of linear equations has infinite solutions
$\therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\Rightarrow \frac{2}{a+b}=\frac{-3}{-(a+b-3)}=\frac{-7}{-(4a+b)}$
$\Rightarrow \frac{2}{a+b}=\frac{3}{(a+b-3)}$
$\Rightarrow 2a+2b-6=3a+3b$
$\Rightarrow b=-a-6$
Again, we have
$\Rightarrow \frac{3}{(a+b-3)}=\frac{7}{(4a+b)}$
$\Rightarrow 12a+3b=7a+7b-21$
$\Rightarrow 5a-4b=-21$
Putting $b=-a-6$
$\Rightarrow 5a-4(-a-6)=-21\Rightarrow 9a=-45$
$\Rightarrow a=-5$
Putting $a=-5inb=-a-6$
$\Rightarrow b=-(-5)-6\Rightarrow b=-1$
$\Rightarrow a=-5,b=-1$