1 of 1

1

(ii) For which value of k will the following pair of linear equations have no solution ?

Soln.

3x

+

y

=

1

(2k

–

1)

x

+

(k

+

1)

y

=

2k

+

1

Comparing equation

(i)

with

a₁x

+

b₁y

+

c₁

=

0

and equation (ii) with

a₂x

+

b₂y

+

c₂

=

0

3x

+

y

–

1 = 0

(2k

–

1)

x

+

(k

–

1)

y

=

2k

+

1

... (i)

(2k

–

1)

x

+

(k

–

1)

y

=

–

0

(2k

+

1)

... (ii)

We get

a₁

=

3

a₂

=

b₁

=

1

b₂

=

c₁

=

–1

c₂

=

2k

–

1

k

–

1

–

(2k

+

1)

a₁

a₂

∴

=

3

2k

–

1

... (iii)

b₁

b₂

=

1

k

–

1

... (iv)

c₁

c₂

=

-1

–

(2k

+

1)

... (v)

For no solution

a₁

a₂

=

b₁

b₂

≠

c₁

c₂

3

2k

–

1

=

1

k

–

1

∴

∴

3k

–

3

2k

–

1

=

∴

k

2

=

3x

+

y

=

1

3(k – 1) = 3k - 3

1(2k – 1) = 2k - 1

3k – 2k

k = 2

= 3 - 1