Question

For what values of $a$ and $b$, the following system of linear equations have an infinite number of solutions
$2x+3y=7$
$(a-b)x+(a+b)y=3a+b-2$

Answer 1

Consider the given equations.

$2x+3y=7$ .....(1)

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

Compare equation (1) with $a_{1}x+b_{1}y+c_1=0$.

$a_1=2,b_1=3,c_1=-7$

Compare equation (2) with$a_{2}x+b_{2}y+c_2=0$.

$a_2=a-b,b_2=a+b,c_2=-(3a+b-2)$

It is given that equation (1) and (2) have infinite number of solutions, then

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

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

Now, $\frac{2}{a-b}=\frac{3}{a+b}$

$2a+2b=3a-3b$

$a=5b$

Now,

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

$9a+3b-6=7a+7b$

Put $a=5b$ in the above result. We get,

$a=5$, $b=1$ .