Question 41154
2x + (k-1)y  =6  ........eq1
3x + (2k+1)y =9 ........eq2


Multiply eq1 by 3 and eq2 by 2 and subtract

 6x + 3(k-1)y  =  18
-6x - 2(2k+1)y = -18


3(k-1)y - 2(2k+1)y  = 0 
y(3(k-1) - 2(2k+1)) = 0
3(k-1) - 2(2k+1) = 0 
3k - 3 - 4k - 2 = 0
-k -5 = 0
-k = 5
k = -5