If the system of equations

3x + y = 1

(2k – 1)x + (k – 1)y = 2k + 1

is inconsistent, then k =
A. 1

B. 0

C. – 1

D. 2

Given:

Equation 1: 3x + y = 1

Equation 2: (2k – 1)x + (k – 1)y = (2k + 1)

Both the equations are in the form of :

a₁x + b₁y = c₁ & a₂x + b₂y = c₂ where

a₁ & a₂ are the coefficients of x

b₁ & b₂ are the coefficients of y

c₁ & c₂ are the constants

For the system of linear equations to have no solutions we must have

………(i)

According to the problem:

a₁ = 3

a₂ = (2k – 1)

b₁ = 1

b₂ = (k – 1)

c₁ = 1

c₂ = (2k + 1)

Putting the above values in equation (i) and solving we get:

⇒ 3 (k – 1) = 2k – 1 ⇒ 3k – 3 = 2k – 1 ⇒ k = 3 – 1 ⇒ k = 2

Therefore = ⇒ =

Putting the value of k we calculate

After comparing the ratio we find

So the given system of equations are inconsistent.

The value of k for which the system of equations is inconsistent is k = 2