Q22 of 186 Page 3

For what value of a, the following system of linear equations has no solutions:

(3a + 1) x + 3y – 2 = 0, (a2 + 1) x + (a – 2) y – 5 = 0

Given, pair of equations


(3a + 1)x + 3y – 2 = 0


and (a2 + 1)x + (a – 2)y – 5 = 0


On comparing the given equation with standard form i.e. a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0, we get


a1 = 3a + 1, b1 = 3 and c1 = – 2


and a2 = a2 + 1, b2 = a – 2 and c2 = – 5


For no solutions,



∴


On taking I and II terms, we get



⇒ (3a + 1)(a – 2) = 3(a2 + 1)


⇒ 3a2 – 6a + a – 2 = 3a2 + 3


⇒ – 5a = 2 + 3


⇒ a = – 1


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