find the value of k for which the equation has no solution kx+3y=3,12x+ky=6​

2 thoughts on “find the value of k for which the equation has no solution kx+3y=3,12x+ky=6​”

1. Given Equation

   kx + 3y = 3 (i)

   12 + ky = 6 (ii)

   To Find the value of k

   We Can Write as

   kx + 3y -3= 0 (i)

   12 + ky-6 = 0 (ii)

   For No Solution

   a₁/a₂ = b₁/b₂ ≠ c₁/c₂

   Now Compare with

   a₁x + b₁y + c₁ = 0

   a₂x + b₂y + c₂ = 0

   We Get

   a₁ = k , b₁ = 3 and c₁= -3

   a₂ = 12 , b₂ = k and c₂ = -6

   Now Put the value value

   a₁/a₂ = b₁/b₂

   k/12 = 3/k

   k² =36

   k = ±6

   Answer

   k = ±6

   Reply

2. Given:

   Two sets of equations are given:

   • kx+3y = 3
   • 12x+ky = 6

   To find:

   • Value of k for which the equation has no solution.

   Solution:

   ➻ kx+3y = 3

   ➻ kx+3y-3 = 0

   where,

   • [tex]\tt{a_1 = k}[/tex]
   • [tex]\tt{b_1 = 3}[/tex]
   • [tex]\tt{c_1 = -3}[/tex]

   Also,

   ➻ 12x+ky = 6

   ➻ 12x+ky-6 = 0

   where,

   • [tex]\tt{a_2 = 12}[/tex]
   • [tex]\tt{b_2 = k}[/tex]
   • [tex]\tt{c_2 = -6}[/tex]

   We know that, if the system of equations has no solution, it is of the form:

   [tex]\tt{\longrightarrow \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \cancel{=} \dfrac{c_1}{c_2}}[/tex]

   Let us substitute the values:

   [tex]\tt{\longrightarrow \dfrac{k}{12} = \dfrac{3}{k} \cancel{=} \dfrac{-3}{-6}}[/tex]

   [tex]\tt{\longrightarrow \dfrac{k}{12} = \dfrac{3}{k}\:and\:\dfrac{3}{k} \cancel{=} \dfrac{-3}{-6}}[/tex]

   By cross multiplication, we get:

   [tex]\tt{\longrightarrow k^2 = 36 \: and \: k \cancel{=} 6}[/tex]

   [tex]\tt{\longrightarrow k = \pm 6 \: and \: k \cancel{=} 6}[/tex]

   Hence, the given system of equations has no solution if k = -6.

   Reply