Question

Write the value of k for which the system of equations x+y-4=0 and 2x +ky-3=0 has no solution.

Answer 1

Given:

The system of questions is

$x+y-4=0$

$2x+ky-3=0$

To find:

The value of k for which the given system of equation has no solution.

Solution:

On comparing the given equations with $ax+by+c=0$, we get

$a_1=1,b_1=1,c_1=-4$

$a_2=2,b_2=k,c_2=-3$

The giver system of equations has no solution, it means both equation represent parallel lines. So,

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}$

Putting the values, we get

$\dfrac{1}{2}=\dfrac{1}{k}\neq \dfrac{-4}{-3}$

Now,

$\dfrac{1}{2}=\dfrac{1}{k}$

On cross multiplication, we get

$k=2$

Therefore, the value of k is 2 for which the given system of equations has no solution.

Answer 2

Step-by-step explanation:

Given:

The system of questions is

x+y-4=0x+y−4=0

2x+ky-3=02x+ky−3=0

To find:

The value of k for which the given system of equation has no solution.

Solution:

On comparing the given equations with ax+by+c=0ax+by+c=0 , we get

a_1=1,b_1=1,c_1=-4a

1

=1,b

1

=1,c

1

=−4

a_2=2,b_2=k,c_2=-3a

2

=2,b

2

=k,c

2

=−3

The giver system of equations has no solution, it means both equation represent parallel lines. So,

\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}

a

2

a

1

=

b

2

b

1



=

c

2

c

1

Putting the values, we get

\dfrac{1}{2}=\dfrac{1}{k}\neq \dfrac{-4}{-3}

2

1

=

k

1



=

−3

−4

Now,

k=2

\dfrac{1}{2}=\dfrac{1}{k}

2

1

=

k

1

On cross multiplication, we get

k=2k=2

Therefore, the value of k is 2 for which the given system of equations has no solution.