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(A) The given pair of linear equations is:$x + 2y - 1 = 0$ .....$(i)$$(a - b)x + (a + b)y - (a + b - 2) = 0$ .....$(ii)$Comparing with $a_1x + b_1y +

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$a = 3, b = 1$

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(A) The given pair of linear equations is:$x + 2y - 1 = 0$ .....$(i)$$(a - b)x + (a + b)y - (a + b - 2) = 0$ .....$(ii)$Comparing with $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$,we get:$a_1 = 1, b_1 = 2, c_1 = -1$$a_2 = (a - b), b_2 = (a + b), c_2 = -(a + b - 2)$For infinitely many solutions,the condition is:$\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}{-(a + b - 2)}$Taking the first two parts:$\frac{1}{a - b} = \frac{2}{a + b} \Rightarrow a + b = 2a - 2b \Rightarrow a = 3b$ .....$(iii)$Taking the last two parts:$\frac{2}{a + b} = \frac{1}{a + b - 2} \Rightarrow 2a + 2b - 4 = a + b \Rightarrow a + b = 4$ .....$(iv)$Substituting the value of $a$ from $(iii)$ into $(iv)$:$3b + b = 4 \Rightarrow 4b = 4 \Rightarrow b = 1$Substituting $b = 1$ into $(iii)$:$a = 3(1) = 3$Thus,for $a = 3$ and $b = 1$,the given pair of equations has infinitely many solutions.

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

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