Question

If the pair of equations kx-3y = 5 and 2x - y = 10 has no solution, then the value of k is​

Answer 1

Answer: For the two lines a  

1

​

x+b  

1

​

y−c  

1

​

=0 anda  

2

​

x+b  

2

​

y−c  

2

​

=0 to have infinite solutions,

a  

2

​

 

a  

1

​

 

​

=  

b  

2

​

 

b  

1

​

 

​

=  

c  

2

​

 

c  

1

​

 

​

 

Here, the lines are 2x+3y−5=0 and 4x+ky−10=0

Thus,  

4

2

​

=  

3

k

​

=  

10

5

​

 

⇒k=6

Answer 2

If the pair of equations kx - 3y = 5 and 2x - y = 10 has no solution, then the value of k = 6

Given :

The pair of equations kx - 3y = 5 and 2x - y = 10

To find :

The value of k for which the pair of equations has no solution

Concept :

For the given two linear equations

$\displaystyle \sf{ a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0}$

Consistent :

One of the Below two condition is satisfied

1. Unique solution :

$\displaystyle \sf{ \: \frac{a_1}{a_2} \ne \frac{b_1}{b_2} }$

2. Infinite number of solutions :

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \: \frac{c_1}{c_2}}$

Inconsistent :

No solution

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \: \frac{c_1}{c_2}}$

Solution :

Step 1 of 2 :

Write down the given equations

Here the given pair of equations are

kx - 3y = 5 - - - - - (1)

2x - y = 10 - - - - - - (2)

Step 2 of 2 :

Find the value of k

Comparing with the equation

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we get

a₁ = k , b₁ = - 3 , c₁ = - 5

a₂ = 2 , b₂ = - 1 , c₂ = - 10

Since the pair of equations have no solution

Thus we get

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \: \frac{c_1}{c_2}}$

$\displaystyle \sf{ \implies \frac{k}{2} = \frac{ - 3}{ - 1} \ne \: \frac{ - 5}{ - 10}}$

$\displaystyle \sf{ \implies \frac{k}{2} = 3}$

$\displaystyle \sf{ \implies k = 6}$

Hence the required value of k = 6

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