Question

i) For which value of k will the following pair of linear equations have no solution
3x+y=1
(2x - 1)x+(k-1)y=2k+1 by substitution method or elimination method ​

Answer 1

The given pair of equations are:

3x+y=1...(1)

(2k−1)x+(k−1)y=2k+1..(2)

Now rearranging eq1 and eq2 will get

3x+y−1=0...(3)

(2k−1)x+(k−1)y−(2k+1)=0..(4)

Now compare with

a

1

=3,b

1

=1,c

1

=−1

a

2

=2k−1,b

2

=k−1,c

2

=−(2k+1)

Now we get

a

2

a

1

=

2k−1

3

,

b

2

b

1

=

k−1

1

,

c

2

c

1

=

−(2k+1)

−1

Now will take

a

2

a

1

=

b

2

b

1

⇒

2k−1

3

=

k−1

1

⇒3k−3=2k−1

⇒3k−2k=−1+3

⇒k=2

Hence k=2 is the value.