Question

Find the value of k the following pair of equation has no solution x+2y=3,(k-1)x+(k+1)y=k+2

Answer 1

These 2 simultaneous linear equations will have no solutions if, 

A / a = B / b 

i.e., the ratio of the x-coefficients is equal to the ratio of y-coefficients. 

x + 2y = 3(k-1)x+(k+1)y = k+2 

First equation: x + 2y = k + 2 
Or, 1x + 2y - (k+2) = 0 [A = 1, B = 2, C = -(k+2)] 

Second equation: 3(k-1)x+(k+1)y = k+2 
Or, (3k-3)x + (k+1)y - (k+2) = 0 [a = 3k-3, b = k+1, c = -(k+2)] 

Required condition: 

A / a = B / b 

Or, 1 / (3k-3) = 2 / (k+1) 
Or, 2 (3k-3) = 1 (k+1) [By cross-multiplication] 
Or, 6k - 6 = k + 1 
Or, 6k - k = 1 + 6 
Or, 5k = 7 

k = 5/7

Answer 2

Since these equationa have no solutions:
a1/a2 =b1/b2¥c1/c2 i.e parallel line
a1=1 b1=2 c1=-3
a2=k-1 b2=k+1 c2=-(k+2)
1/k-1 = 2/k+1 ¥ -3/-k-2
k+1 =2(k-1)
k+1 = 2k-2
2k-k=2+1
k=3