**Question:**

For which values of a and b will the following pair of linear equations has infinitely many

solutions?

x + 2y = 1

(a -b)x+(a + b)y = a + b – 2

**Solution:**

Given pair of linear equations are

x+2y=1

(a-b)x+(a+b)y=a+b-2

on comparing with ax+by+c=0,we get

$a_{1}=1, b_{4}=2$ and $c_{4}=-1 \quad$ [from Eq. (i)]

$a_{2}=(a-b), b_{2}=(a+b) \quad$ [from Eq. (ii)]

and $\quad c_{2}=-(a+b-2)$

For infinitely many solutions of the the pairs of linear equations,

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \quad \frac{1}{a-b}=\frac{2}{a+b}=\frac{-1}{-(a+b-2)}$

Taking first two parts,

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

$\Rightarrow \quad a+b=2 a-2 b$

$\Rightarrow \quad 2 a-a=2 b+b$

$\Rightarrow \quad a=3 b$ \ldots. (iii)

Taking last two parts,

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

$\Rightarrow \quad 2 a+2 b-4=a+b$

$\Rightarrow \quad a+b=4$ .....(iv)

Now, put the value of a from $\mathrm{Eq}$. (iii) in Eq. (iv), we get

$3 b+b=4$

$\Rightarrow \quad 4 b=4$

$\Rightarrow \quad b=1$

Put the value of $b$ in Eq. (iii), we get

$a=3 \times 1$

$\Rightarrow \quad a=3$

So, the values $(a, b)=(3,1)$ satisfies all the parts. Hence, required values of $a$ and $b$ are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions.