Question

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

• A. $a=-5,b=-1$
• B. $a=1,b=5$
• C. $a=-1,b=-5$
• D. $a=5,b=1$

Answer 1

The correct option is A $a=-5,b=-1$

The given system of equations can be written as

$2x-3y=7$

$\Rightarrow$ $2x-3y-7=0$ ----- ( 1 )

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

$\Rightarrow$ $(a+b)x-(a+b-3)y-(4a+b)=0$ ------ ( 2 )

Comparing given equations with following forms,

$a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$

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

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

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

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

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

$\Rightarrow$ $2(4a+b)=7(a+b)$ and $3(4a+b)=7(a+b-3)$

$\Rightarrow$ $8a+2b=7a+7b$ and $12a+3b=7a+7b-21$

$\therefore$ $a=5b$ ---- ( 3 )

and $5a=4b-21$ ----- ( 4 )

On substituting $a=5b$ in ( 4 ), we get,

$5\left(5b\right)=4b-21$

$\Rightarrow$ $25b=4b-21$

$\Rightarrow$ $21b=-21$

$\therefore$ $b=-1$

On substituting $b=-1$ in ( 3 ), we get

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

$\therefore$ $a=-5$ and $b=-1$