Question

find the value of k for which the following pair of linear equation has no solution 3x+y=2 (2k-1)x+(k-1)y=2k-1​

Answer 1

Step-by-step explanation:

Given :-

3x+y=2

(2k-1)x+(k-1)y=2k-1

To find :-

Find the value of k for which the following pair of linear equations has no solution ?

Solution :-

Given pair of linear equations in two variables are 3x+y=2

=> 3x+y-2 = 0

On comparing with a1x+b1y+c1 = 0 then

a1 = 3

b1 = 1

c1 = -2

and

(2k-1)x+(k-1)y=2k-1

=> (2k-1)x+(k-1)y-(2k-1) = 0

On comparing with a2x+b2y+c2 = 0 then

a2 = 2k-1

b2 = k-1

c2 = -(2k-1)

Given that

Given equations has no solution

=> a1/a2 = b1/b2

=> 3/(2k-1) = 1/(k-1)

On applying cross multiplication then

=> 3(k-1) = 1(2k-1)

=> 3k-3 = 2k-1

=> 3k-2k = -1+3

=> k = 2

Therefore, k = 2

Answer :-

The value of k for the given problem is 2

Used formulae:-

Let a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 are the pair of linear equations in two variables,

if a1/a2 = b1/b2 then they have no solution .

Points to know:-

Let a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 are the pair of linear equations in two variables,

• If a1/a2 ≠ b1/b2 then they are Consistent and independent lines or Intersecting lines with a unique solution .
• If a1/a2 = b1/b2 then they are Inconsistent lines or Parallel lines with no solution .
• If a1/a2 = b1/b2 = c1/c2 then they are Consistent and dependent lines or Coincident lines with infinitely number of many solutions .