Question -
Answer -
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
Solution:
(i) 3y + 2x -7 =0
(a + b)y + (a-b)y – (3a + b -2) = 0
a1/a2 = 2/(a-b) , b1/b2 = 3/(a+b) , c1/c2 = -7/-(3a + b -2)
For infinitely many solutions,
a1/a2 = b1/b2 = c1/c2
Thus 2/(a-b) = 7/(3a+b– 2)
6a + 2b – 4 = 7a – 7b
a – 9b = -4 ……………………………….(i)
2/(a-b) = 3/(a+b)
2a + 2b = 3a – 3b
a – 5b = 0 ……………………………….….(ii)
Subtracting (i) from (ii), we get
4b = 4
b =1
Substituting this eq. in (ii), we get
a -5 x 1= 0
a = 5
Thus at a = 5 and b = 1 the given equations will have infinite solutions.
(ii) 3x + y -1 = 0
(2k -1)x + (k-1)y – 2k -1 = 0
a1/a2 = 3/(2k -1) , b1/b2 = 1/(k-1), c1/c2 = -1/(-2k -1) = 1/( 2k +1)
For no solutions
a1/a2 = b1/b2 ≠ c1/c2
3/(2k-1) = 1/(k -1) ≠ 1/(2k +1)
3/(2k –1) = 1/(k -1)
3k -3 = 2k -1
k =2
Therefore, for k = 2 the given pair of linear equations will have no solution.