Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$(2 a - 1) x + 3 y = 5, 3 x + (b - 1) y = 2 .$

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Solution

The given system of equations:
(2a − 1)x + 3y = 5
⇒ (2a − 1)x + 3y − 5 = 0 ....(i)
And, 3x + (b − 1)y = 2
⇒ 3x + (b − 1)y − 2 = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = (2a − 1), b₁= 3, c₁ = −5 and a₂ = 3, b₂= (b − 1), c₂ = −2
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\frac{(2 a - 1)}{3} = \frac{3}{(b - 1)} = \frac{- 5}{- 2}$
$\Rightarrow \frac{(2 a - 1)}{3} = \frac{3}{(b - 1)} = \frac{5}{2}$
$\Rightarrow \frac{(2 a - 1)}{3} = \frac{5}{2} and \frac{3}{(b - 1)} = \frac{5}{2}$
⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)
⇒ 4a − 2 = 15 and 6 = 5b − 5
⇒ 4a = 17 and 5b = 11
$\Rightarrow a = \frac{17}{4} and b = \frac{11}{5}$
∴ a = $\frac{17}{4}$ and b = $\frac{11}{5}$

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