Find the value of k for which the following system of equations has no solution 2x+ky=11 5x-7y=5

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Find the value of k for which the following system of equations has no solution 2x+ky=11 5x-7y=5

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Find the value of k for which the following system of equations has no solution 2x+ky=11 5x-7y=5

Find the value of k for which each of the following system of equations have no solution :2 x+k y=115 x 7 y=5

The given system of equation is

2x + ky - 11 =0

5x − 7y - 5 = 0

The system of equation is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where, `a_1 = 2,b_1 = k, c_1 = -11`

And `a_2 = 5, b_2 = -7, c_2 = -5`

For a unique solution, we must have

`a_1/a_2 - b_1/b_2 != c_1/c_2`

`=> 2/5 = k/(-7) != (-11)/(-5)`

Now

`2/5 = k/(-7)`

`=> 2 xx (-7) = 5k`

`=> 5k = -14`

`=> k = (-14)/5`

Cleary for `(-14)/5` we have `k/(-7) != (-11)/(-5)`

Hence, the given system of equation will have no solution if `k = (-14)/5`


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Find the value of k for which the following system of equations has a unique solution:

kx + 3y = 3

12x + ky = 6

kx + 3y = 3 

12x + ky = 6

For no solution `a_1/a_2 - b_1/b_2 != c_1/c_2`

`=> k/12 = 2/k != 3/6`

`k/12 = 3/k`

`k^2 = 36`

`k = +- 6` i.e k = 6, -6

Also

`3/k != 3/6`

`(3 xx 6)/3 != k`

`k != 6`

k = -6 satisfies both the condition

Hence k = -6

Concept: Pair of Linear Equations in Two Variables

  Is there an error in this question or solution?


Page 3

The given system of equation may be written as

4x + 6y - 11 = 0

2x + ky - 7 = 0

The system of equation is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where `a_1 = 4, b_1 = 6, c_1 = -11`

And `a_2 = 2, b_2 = k, c_2 = -7`

For a unique solution, we must have

`a_1/a_2 = b_1/b_2 != c_1/c_2`

Now

`a_1/a_2 = b_1/b_2`

`=> 4/2 = 6/k`

`=> 4k = 12`

`=> k = 12/4 = 3`

Clearly, for this value of k, we have

`a_1/a_2 = b_1/b_2 != c_1/c_2

Hence, the given system of equation is inconsistent, when k = 3


Page 4

The given system of the equation may be written as

αx + 3y -α - 3 = 0

12x + αy - α = 0

The system of equation is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where `a_1 = α, b_1 = 3, c_1 = -(α - 3)`

And `a_2 = 12, b_2 = α, c_2 = -α`

For a unique solution, we must have


`a_1/a_2 - b_1/b_2 != c_1/c_2`

`=> α/12 = 3/α != (-(α - 3))/(-α)`

Now,

`3/α != (-(α - 3))/(-α)`

`=> 3/α != (α - 3)/α`

`=> 3 != α - 3`

`=> 3 = 3 != α`

`=> 6 != α`

`=> α != 6`

And

`α/12 = 3/α`

`=> α^2 = 36`

`=> α +- 6`

`=> α = -6`       [∵ `α != 6`]

Hence, the given system of equation will have no solution if α = -6