1.

Determine the value of k so that the following linear equations have nosolution : (3k + 1)x + 3y - 2 = 0 and (k^2 + 1)x + (k - 2)y – 5 = 0.​

Answer»

Step-by-step EXPLANATION:

The given system of equations is

(3k+1)x+3y−2=0

(k

2

+1)x+(k−2)y−5=0

This is of the form a

1

x+b

1

y+C

1

=0

a

2

x+b

2

y+c

2

=0,

where, a

1

=3k+1,b

1

=3,c

1

=−2

and a

1

=k

2

+1,b

2

=k−2,c

2

=−5

For no SOLUTION, we must have

a

2

a

1

=

b

2

b

1

=

c

2

c

1

The given system of equations will have no solution, if

k

2

+1

3k+1

=

k−2

3

=

−5

−2

⇒

k

2

+1

3k+1

=

k−2

3

and

k−2

3

=

5

2

Now,

k

2

+1

3k+1

=

k−2

3

⇒(3k+1)(k−2)=3(k

2

+1)

⇒3k

2

−5k−2=3k

2

+3

⇒−5k−2=3

⇒−5k=5

⇒k=−1

Clearly,

k−2

3

=

5

2

for k=−1

Hence, the given system of equations will have no solution for k=−1.


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