Question

Find the values of a and b for which the system of linear equations has an infinite number of solutions:
(a – 1) x + 3y = 2, 6x + (1 – 2b)y = 6

Answer 1

The given system of equations can be written as
(a – 1) x + 3y = 2
⇒(a – 1) x + 3y – 2 = 0           ….(i)
and 6x + (1 – 2b)y = 6
⇒6x + (1 – 2b)y – 6 = 0         ….(ii)
These equations are of the following form:
`a_1x+b_1y+c_1 = 0`
`a_2x+b_2y+c_2 = 0`
where, `a_1 = (a – 1), b_1= 3, c_1= -2 and a_2 = 6, b_2 = (1 – 2b), c_2= -6`
For an infinite number of solutions, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`
`⇒ (a−1)/6 = 3/((1−2b)) = (−2)/(−6)`
`⇒ (a−1)/6 = 3/((1−2b)) = 1/3`
`⇒ (a−1)/6 = 1/3 and 3/((1−2b)) = 1/3`
⇒ 3a – 3 = 6 and 9 = 1 – 2b
⇒ 3a = 9 and 2b = -8
⇒ a = 3 and b = -4
∴ a = 3 and b = -4