Find the value of k for which the simultaneous equations below have no solution.?

Question

2x-ky = 2

and

3x+(k+1)y = 4

2x-ky = 2

and

3x+(k+1)y = 4


can anyone solve it without using parallel lines?

Answer 1

No solution:

2/3 = -k/(k + 1) but not equal to 1/2

2/3 = -k/(k + 1)
2(k + 1) = -3k
2k + 2 = -3k
5k + 2 = 0
k = -2/5

-2/5 does not equal 1/2

So, if k = -2/5, this system has no solution


There is a simple theory you can apply:

Consider the system:

a1x + b1y + c1 = 0
a2x + b2y + c2 = 0

If the system has a unique solution:
a1/a2 does not equal b1/b2 does not equal c1/c2

If the system has no solution:
a1/a2 = b1/b2 but does not equal c1/c2

If the system has infinite solutions:
a1/a2 = b1/b2 = c1/c2

Answer 2

These 2 lines has no solution when both lines are parallel.

2x-ky = 2
or equal to ky = 2x-2
then m1 = 2/k

3x+(k+1)y = 4
or equal to (k+1)y = 4-3x
then m2 = (-3)/(k+1)

If m1 = m2
2/k = (-3)/(k+1)
2k+2 = -3k
k = 2/5

Answer 3

k= -2/5

because,

2x - ky =2
3x + (k-1)= 4

here a1/a2= 2/3, b1/b2=-k/k+1, c1/c2=2/4=1/2

for no sol., v must hav,
a1/a2=b1/b2 !=(is not =to) c1/c2

so, 2/3=-k/k+1!=(is not =to) 1/2

2k+2 =-3k
5k = -2
k = -2/5
also, -k !=(is not =to) 1/2
-2k != k +1
-3k != 1
k!= -1/3

This is the complete calculation

Answer 4

k=2(1-y)/5yprovided y=0