Question

find the value of k if the pair of equations 2x+3y-9 =0 and 4x+ky-18=0 has infinitely many solutions​

Answer 1

Answer:

The required value of k is 6.

Step-by-step explanation:

Given :

The pair of equations 2x + 3y - 9 = 0 and 4x + ky - 18 = 0 has infinitely many solutions​

To find :

the value of k

Solution :

  Comparing the given two equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get

a₁ = 2 , b₁ = 3 , c₁ = -9

a₂ = 4 , b₂ = k , c₂ = -18

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 has infinitely many solutions when

  $\boxed{\tt \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}}$

Substitute the values,

$\sf \dfrac{2}{4}=\dfrac{3}{k}=\dfrac{-9}{-18} \\\\ \sf \dfrac{1}{2}=\dfrac{3}{k}=\dfrac{1}{2} \\\\ \implies \sf \dfrac{3}{k}=\dfrac{1}{2} \\\\ \implies \sf k=3 \times 2 \\\\ \implies \sf k=6$

The value of k is 6

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Know more :

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 has

1) no solution when   $\tt \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

2) infinite solutions when   $\tt \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

3) unique solution when   $\tt \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$