Question

Find the value of k for which the following system of equations has a unique solution:
kx+2y=53x+y=1

Answer 1

Given pair of system of equations  :  

kx + 2y = 5

3x + y = 1

 

 

The given pair of  linear equation can be written as :

kx + 2y - 5 = 0………(1)

3x + y - 1 = 0…………(2)

 

On comparing with General form of a pair of linear equations  

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

 a1 = k, b1 = 2, c = - 5

a2 = 3 , b2 = 1 , c = -1

We have ,  

a1/a2 = k/3 ,  b1/b2 = 2/1 & c1/c2 = -5/-1

 

Given: A pair of linear equations has a unique solution, if a1/a2 ≠ b1/b2

k/3 ≠ 2/1

1 × k ≠ 3 × 2

k ≠ 6

Hence, given lines have unique solution for all real values of k, except 6.

 Hope this answer will help you…

 

Some more questions from this chapter :  

Find the value of k for which pair of linear equations 3x + 2y = –5 and x - ky = 2 has a unique solution.

https://brainly.in/question/2798670

 

For what value of p the pair of linear equations (p+ 2)x - (2p + 1)y = 3(2p -1) and 2x- 3y = 7 has a unique solution.

https://brainly.in/question/2810701

Answer 2

$\huge\mathcal{Answer:}$

$kx + 2y - 5 = 0$

$3x + y - 1 = 0$

$a _{1}x + b_{1}y + c_{1} = 0$

$a _{2}x + b _{2}y + c _{2} = 0$

$a _{1} = k \: \: b _{1} = 2 \: c _{1} = - 5$

$a_{2} = 3 \: \: b _{2} = 1 \: c _{2} = - 1$

$\frac{a _{1} }{a _{2}} ≠ \frac{b _{1} }{b _{2} }$

$\frac{k}{3} ≠ \frac{2}{1}$

$k≠6$

Answer is k=6