FIND THE MISSING VALUE IF THE SYSTEM HAS NO SOLUTION

Find the value of k, so that the following system of equations has no solution:

Problem 1 :

3x + y = 1; (2k - 1)x + (k - 1)y = (2k - 1)

Solution :

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

3/(2k - 1) = 1/(k-1) ≠ 1/(2k - 1)

3(k - 1) = 1(2k - 1)

3k - 3 = 2k - 1

3k - 2k = -1 + 3

 k = 2

Hence, the required value of k is 2.

Problem 2 :

3x + y = 1; (2k - 1)x + (k - 1)y = (2k + 1)

Solution :

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

3/(2k - 1) = 1/(k-1) ≠ 1/(2k + 1)

3(k - 1) = 1(2k - 1)

3k - 3 = 2k - 1

3k - 2k = -1 + 3

 k = 2

Hence, the required value of k is 2.

Problem 3 :

x - 2y = 3; 3x + ky = 1

Solution :

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

1/3 = -2/k ≠ -3/-1

1/3 = -2/k

k = -6

Hence, the required value of k is -6.

Problem 4 :

x + 2y = 5; 3x + ky + 15 = 0

Solution :

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

1/3 = 2/k ≠ -5/15

1/3 = 2/k and 2/k ≠ -5/15

k = 6 and k ≠ -6

Hence, the required value of k is 6.

Problem 5 :

kx + 2y = 5; 3x - 4y = 10

Solution :

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

k/3 = 2/-4 ≠ -5/-10

k/3 = 2/-4 and k/3 ≠ 1/2

k = -3/2 and k ≠ 3/2

Hence, the required value of k is -3/2.

Problem 6 :

x + 2y = 3; 5x + ky + 7 = 0

Solution :

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

1/5 = 2/k ≠ -3/7

1/5 = 2/k and 2/k ≠ -3/7

k = 10 and k ≠ 14/-3

Hence, the required value of k is 10.

Problem 7 :

8x + 5y = 9; kx + 10y = 15

Solution :

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

8/k = 5/10 ≠ -9/-15

8/k = 1/2 ≠ 3/5

8/k = 1/2 and 8/k ≠ 3/5

k = 16 and k ≠ 40/3

Hence, the required value of k is 16.