Let the system of linear equations be
a_{1}x + b_{1}y + c_{1} = 0
a_{2}x + b_{2}y + c_{2} = 0
If the system has unique solution, then
a_{1}/a_{2} ≠ b_{1}/b_{2}
If the system has infinitely many solution, then
a_{1}/a_{2} = b_{1}/b_{2 }= c_{1}/c_{2}
If the system has no solution, then
a_{1}/a_{2} = b_{1}/b_{2 }≠ c_{1}/c_{2}
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 :
a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
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 :
a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
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 :
a_{1}/a_{2} = b_{1}/b_{2 }≠ c_{1}/c_{2}
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 :
a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
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 :
a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
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 :
a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
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 :
a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}
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.
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