Question

For which values of $a$ and $b$, will the following pair of linear equations have infinitely many solutions?
$x+2y=1$; $(a-b)x+(a+b)y=2$

Answer 1

Given equations $x+2y=1⟶\left(1\right)$

$(a-b)x+(a+b)y=2⟶\left(2\right)$.

Comparing $x+2y=1$ with $a_{1}x+b_{1}y+c_1=0$,

and $(a-b)x+(a+b)y=2$ with $a_{2}x+b_{2}y+c_2=0$,

we get, $a_1=1,a_2=a-b,b_1=2,b_2=a+b,c_1=-1,c_2=-2$

Now,

For given system to have infinity no. of solutions,

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

$\frac{1}{a-b}=\frac{2}{a+b}=\frac{1}{2}$

$\therefore \frac{1}{a-b}=\frac{1}{2},\frac{2}{a+b}=\frac{1}{2}$

$a-b=2⟶\left(3\right)a+b=4⟶\left(4\right)$

Solving $\left(3\right)$ and $\left(4\right)$

$2a=6⟹a=3$

$⟹b=1$

The values of $a$ and $b$ are $3$ and $1$.