Question

find the value of k for which the systems of equations x - 2 y = 3 and 3 x + k y = 1 has a unique solution ​

Answer 1

Answer:

For all real values of k , (k≠6) given system of equations have unique solution.

Step-by-step explanation:

Compare given system of equations

x-2y=3=> x-2y-3=0 ---(1)

3x+ky=1=> 3x+ky-1=0 ---(2) with

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

1

x+b

1

y+c

1

=0

a_{2}x+b_{2}y+c_{2}=0,we\:geta

2

x+b

2

y+c

2

=0,weget

\begin{gathered}a_{1}=1,b_{1}=-2,c_{1}=-3\\ a_{2}=3,b_{2}=k,c_{1}=-1\end{gathered}

a

1

=1,b

1

=−2,c

1

=−3

a

2

=3,b

2

=k,c

1

=−1

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

a

2

a

1



=

b

2

b

1

/* Given lines have unique solution*/

\implies \frac{1}{3}≠\frac{-2}{k}⟹

3

1



=

k

−2

\implies 1\times k≠-2 \times 3⟹1×k



=−2×3

\implies k≠-6⟹k



=−6

Therefore,

For all real values of k , (k≠6) given system of equations have unique solution.