Question

find the value of k for which the pair of linear equation 3x+ 5y=3, 6x+y= 8 do not have solution

Answer 1

3x+5y-3=0
6x+y-8=0
First of all there is no k
So I put k in place of coefficient of y in 2 equation
Value of k will be 10

Answer 2

Dear Student,

Answer: k = 10

Solution: Condition for inconsistency or no solution

As we know standard equation in two variable is represented as

a1 x +b1 y+ c1 =0

a2 x +b2 y+ c2 =0

$\frac{a1}{a2} = \frac{b1}{b2} \neq\frac{c1}{c2}$

3x+ 5y-3=0, a1=3 , b1 = 5 , c1 = -3

6x+ky- 8 =0, a2 = 6 , b2 = k , c2 = -8 ( made correction for solution)

$\frac{3}{6} = \frac{5}{k} \neq\frac{-3}{-8}$

1/2 = 5/k

so, k= 10

Hope it helps you.