(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a - b)x + (a + b)y = 3a + b - 2
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k - 1)x + (k - 1)y = 2k + 1

Explanation

(i) We have, 2x + 3y - 7 = 0
and (a - b)x + (a + b)y - (3a + b - 2) = 0
Here, `a_(1) = 2, b_(1) = 3, c_(1) = - 7, a_(2) = a - b, b_(2) = a + b, c_(2) = - (3a + b - 2)`
For infinite number of solutions, we have
`(2)/(a-b) = (3)/(a+b)=(-7)/(-(3a + b - 2)) (because (a_(1))/(a_(2)) = (b_(1))/(b_(2)) = (c_(1))/(c_(2)))`
`implies (a-b)/(2) = (a+b)/(3) = (3a + b - 2)/(7)`
From first and second, we have
`(a-b)/(2) = (a+b)/(3)`
implies 3a - 3b = 2a + 2b
implies a = 5b ...(1)
From second and third, we have
`(a + b)/(3) = (3a + b - 2)/(7)`
implies `7a + 7b = 9a + 3b - 6`
implies 4b = 2a - 6
implies 2b = a - 3 (Divide by 2) ...(2)
From equations (1) and (2), we get
2b = 5b - 3 implies b = 1
Substituting b = 1 in equation (1), we get
a = 5 `xx` 1 implies a = 5
(ii) We have, 3x + y - 1 = 0 ...(1)
and (2k - 1)x + (k - 1) y - (2k + 1) = 0 ...(2)
Here, `a_(1) = 3, b_(1) = 1, c_(1) = - 1, a_(2) = 2k - 1, b_(2) = k - 1, c_(2) = - (2k + 1)`
`therefore" "(a_(1))/(a_(2)) = (3)/(2k - 1), (b_(1))/(b_(2)) = (1)/(k - 1), (c_(1))/(c_(2)) = (1)/(2k + 1)`
For no solution, `(a_(1))/(a_(2)) = (b_(1))/(b_(2)) ne (c_(1))/(c_(2))" i.e., "(3)/(2k - 1) = (1)/(k - 1) ne (1)/(2k + 1)`
implies 3(k - 1) = 2k - 1 and k - 1 `ne` 2k + 1
implies 3k - 3 = 2k - 1 and `k ne - 2`
implies k = 2 and `k ne - 2`
Hence, given system has no solution when k = 2.