Question

For what value of k, the pair of linear equations 3x+y=3 and 6x + ky = 8

does not have a solutio



​

Answer 1

Answer:

The required value of k is 2

Step-by-step explanation:

Given :

the pair of linear equations

• 3x + y = 3

• 6x + ky = 8

To find :

the value of k for which the given pair of equations does not have a solution

Solution :

⇒ 3x + y = 3

3x + y - 3 = 0

⇒ 6x + ky = 8

6x + ky - 8 = 0

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

a₁ = 3 , b₁ = 1 , c₁ = -3

a₂ = 6 , b₂ = k , c₂ = -8

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 do not have a solution when

    $\sf \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

Substitute the values,

  $\rm \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} \\\\ \rm \dfrac{3}{6}=\dfrac{1}{k} \neq \dfrac{-3}{-8} \\\\ \rm \dfrac{3}{3 \times 2}=\dfrac{1}{k} \neq \dfrac{3}{8} \\\\ \rm \dfrac{1}{2}=\dfrac{1}{k} \\\\ \rm 1 \times k= 1 \times 2 \\\\ \sf k=2$

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

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

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

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

Answer 2

$\large\mathfrak{\blue{Given}}$

The pair of linear equations :

• 3x + y = 3
• 6x + ky = 8

$\large\mathfrak{\blue{To ~find}}$

• The value of k for which the given pair of equations does not have a solution.

$\huge\mathfrak{\color{orange}{Solution :}}$

• 3x + y = 3

$\implies$3x + y - 3 = 0

• 6x + ky = 8

$\implies$6x + ky - 8 = 0

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

a₁ = 3 , b₁ = 1 , c₁ = -3

a₂ = 6 , b₂ = k , c₂ = -8

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 do not have a solution when :

$\sf \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

Substitute the values,

$\implies\rm \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

$\implies\rm \dfrac{3}{6}=\dfrac{1}{k} \neq \dfrac{-3}{-8}$

$\implies\rm \dfrac{3}{3 \times 2}=\dfrac{1}{k} \neq \dfrac{3}{8}$

$\implies\rm \dfrac{1}{2}=\dfrac{1}{k}$

$\implies\rm 1 \times k= 1 \times 2$

$\implies{\red{\sf k=2}}$

∴ The required value of k is 2

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

(1) No solution when $\sf \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

(2) Infinite solutions when  $\sf \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

(3) Inique solution when  $\sf \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$