Question

Determine the value of k so that the following linear equations have a unique solution:
(3k+1)x+3y-2=0
(k²+1)x+(k-2)y-5=0

Answer 1

Answer:

Except -1 for all real values of k given the equations has a unique solution.

Explanation:

Compare given linear pair of equations

(3k+1)x+3y-2 = 0 ----(1)

(k²+1)x+(k-2)y-5=0 ----(2)

with

$a_{1}x+b_{1}y+c_{1}=0$----(2)

$a_{2}x+b_{2}y+c_{2}=0$----(3)

we get ,

$a_{1}=(3x+1), b_{1}=k^{2}+1, c_{1}=-2$

and

$a_{2}=k^{2}+1, b_{2}=k-2, c_{2}=5$

$\frac{a_{1}}{a_{2}}≠\frac{b_{1}}{b_{2}}$

\* Given linear equations have unique solution *\

$\implies \frac{3k+1}{k^{2}+1}≠\frac{3}{k-2}$

$(3k+1)(k-2)≠3\times(k^{2+1)$

=> 3k(k-2)+1(k-2)≠3k²+3

=> 3k²-6k+k-2≠3k²+3

=> 3k²-5k-3k²≠3+2

=> -5k ≠ 5

Divide each term by (-5) , we get

=> k ≠-1

Therefore,

Except -1 for all real values of k given the equations has a unique solution.

••••

Answer 2

Step-by-step explanation:

Answer:

Except -1 for all real values of k given the equations has a unique solution.

Explanation:

Compare given linear pair of equations

(3k+1)x+3y-2 = 0 ----(1)

(k²+1)x+(k-2)y-5=0 ----(2)

with

a_{1}x+b_{1}y+c_{1}=0a

1

x+b

1

y+c

1

=0 ----(2)

a_{2}x+b_{2}y+c_{2}=0a

2

x+b

2

y+c

2

=0 ----(3)

we get ,

a_{1}=(3x+1), b_{1}=k^{2}+1, c_{1}=-2a

1

=(3x+1),b

1

=k

2

+1,c

1

=−2

and

a_{2}=k^{2}+1, b_{2}=k-2, c_{2}=5a

2

=k

2

+1,b

2

=k−2,c

2

=5

\frac{a_{1}}{a_{2}}≠\frac{b_{1}}{b_{2}}

a

2

a

1



=

b

2

b

1

\* Given linear equations have unique solution *\

\implies \frac{3k+1}{k^{2}+1}≠\frac{3}{k-2}⟹

k

2

+1

3k+1



=

k−2

3

(3k+1)(k-2)≠3\times(k^{2+1)

=> 3k(k-2)+1(k-2)≠3k²+3

=> 3k²-6k+k-2≠3k²+3

=> 3k²-5k-3k²≠3+2

=> -5k ≠ 5

Divide each term by (-5) , we get

=> k ≠-1

Therefore,

Except -1 for all real values of k given the equations has a unique solution.

••••