Question

The value of k for system of equation 2x+kyx+3y-4=0 is inconsistent​

Answer 1

Answer:

K= 6

Step-by-step explanation:

The value of k for which the system of equations 2x+ky  x+3y-4=0 are inconsistent is

the question has some typing error

since its given that the system of equations it shows only equation, so split the given equation as

 2x+ky =0

x+3y-4 =0

its also given the system of equation is inconsistent

therefore

a1/a2  =  b1/b2  

2/1   = k/ 3

hence k=6

Answer 2

Given : system of equation 2x+ky , x+3y-4=0

To Find : value of k so that its inconsistent​

Solution:

a₁x+ b₁y=c₁

a₂x+b₂y=c₂

Consistent

if a₁/a₂ ≠ b₁/b₂  ( unique solution)

a₁/a₂ = b₁/b₂ = c₁/c₂  ( infinite solution )

Inconsistent

if  a₁/a₂ = b₁/b₂ ≠  c₁/c₂  ( No solution)

2x+ky = 0

x + 3y  = 4

a₁/a₂ = 2/1 = 2

b₁/b₂ = k/3

c₁/c₂  = 0/4 = 0

inconsistent​ if  a₁/a₂ = b₁/b₂ ≠  c₁/c₂

=> 2 = k/3 ≠ 0

=> k  = 6

value of k is 6  for system of equation 2x+ky ,  x+3y-4=0 to be inconsistent​

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