Question

for what value of k , the pair of linear equation 3x + y=3 6x + ky =8
does not have a solution​

Answer 1

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The equation are

3x + y − 3 = 0     ....(1)

6x + ky − 8 = 0     .....(2)

Here, a1 = 3 , b1 = 1 , c1 = −3

and a2 = 6 , b2 = k , c2 = −8

The   given   system   has   no   solution   if

$\frac{a1}{a2} = \frac{b1}{b2} \: not \: equal \: \: \frac{c1}{c2}$

$\frac{3}{6} = \frac{1}{k}$

$k = \frac{6}{3}$

$k = 2$

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Answer 2

Answer:

k = 2, k ≠ 8/3

Step-by-step explanation:

Given:

A pair of linear equations:

• 3x + y = 3
• 6x + ky = 8

To Find:

The value of k so the pair of equations does not have a solution

Solution:

Here we are given a pair of linear equations.

We have to find the value of k so that the equations do not have a solution.

The equations are:

3x + y = 3

6x + ky = 8

where a₁ = 3, a₂ = 6, b₁ = 1, b₂ = k, c₁ = 3, c₂ = 8

A pair of linear equation has no solution if

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

Substituting the data,

$\sf{\dfrac{3}{6}=\dfrac{1}{k} \neq \dfrac{3}{8}}$

Equating the first part we get,

$\sf{\dfrac{3}{6} =\dfrac{1}{k} }$

3k = 6

 k = 6/3

k = 2

Equating the second part,

$\sf{\dfrac{1}{k}\neq \dfrac{3}{8} }$

3k ≠ 8

k ≠ 8/3

Hence the value of k is 2 and it is not equal to 8/3

$\boxed{\bold{k=2,k\neq \dfrac{8}{3}}}$

Notes:

If the linear equations have a unique solution and is consistent,

$\sf{\dfrac{a_1}{b_1} \neq \dfrac{b_1}{b_2}}$

If the linear equations have infinite number of solutions and is consistent,

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

If the linear equations have no solution and is inconsistent.

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