For which values of \(a\) and \(b\) will the pair of equations \(x + 2y = 1\) and \((a-b)x + (a+b)y = a + b - 2\) have infinitely many solutions?
\(a = 3\) and \(b = 1\).
Write coefficients:
\(a_1 = 1,\ b_1 = 2,\ c_1 = -1\).
\(a_2 = a-b,\ b_2 = a+b,\ c_2 = -(a+b-2)\).
For infinitely many solutions,
\(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}.\)
From the first two ratios:
\(\dfrac{1}{a-b} = \dfrac{2}{a+b}\Rightarrow a+b = 2(a-b)\).
So, \(-a + 3b = 0\Rightarrow a = 3b\).
Match with constants ratio:
\(\dfrac{1}{a-b} = \dfrac{-1}{-a-b+2} = \dfrac{1}{a+b-2}\).
Thus \(a-b = a+b-2\Rightarrow -b = b - 2\Rightarrow b = 1\).
Then \(a = 3b = 3\).