Write the value of k for which the system of equations x+y-4=0 and 2x +ky-3=0 has no solution.
Answers
Given:
The system of questions is
To find:
The value of k for which the given system of equation has no solution.
Solution:
On comparing the given equations with , we get
The giver system of equations has no solution, it means both equation represent parallel lines. So,
Putting the values, we get
Now,
On cross multiplication, we get
Therefore, the value of k is 2 for which the given system of equations has no solution.
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.