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Find the Values of K for Which the System 2x + Ky = 1 3x – 5y = 7 Will Have (I) a Unique Solution, and (Ii) No Solution. is There a Value of K for Which the System Has Infinitely Many Solutions? - CBSE Class 10 - Mathematics

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Question

Find the values of k for which the system
2x + ky = 1
3x – 5y = 7
will have (i) a unique solution, and (ii) no solution. Is there a value of k for which the
system has infinitely many solutions?

Solution

The given system of equation may be written as

2x + ky - 1 = 0

3x – 5y - 7 = 0

It 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 = -1`

And `a_2 = 3, b_2 = -5, c_2 = -7`

1) The given system will have a unique solution, if

`a_1/a_2 != b_1/b_2`

`=> 2/3 != k/(-5)`

`=> -10 != 3k`

`=> 3k != - 10`

`=> k != (-10)/3`

So, the given system of equations will have a unique solution if k = (-10)/3

2) The given system will have no solution, if

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

We have

`a_1/a_2 = b_1/b_2`

`=> 2/3 = k/(-5)`

`=> -10 = 3k`

=> 3k = -10

`=> k = (-10)/3`

We have

`b_1/b_2 = k/(-5) = (-10)/(3 xx -5) = 2/3`

And `c_1/c_2 = (-1)/(-7) = 1/7`

Clearly `b_1/b_2 != c_1/c_2`

So, the given system of equations will have no solution , if `k = (-10)/3`

For the given system to have infinite number of solutions, we must have

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

We have,

`a_1/a_2 = 2/3, b_1/b_2 = k/(-5)`

And `c_1/c_2 = (-1)/(-7) = 1/7`

Clearly `a_1/a_2 != c_1/c_2``

So, whatever be the value of k, we cannot have

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

Hence, there is no value of k, for which the given system of equations has infinitely many solutions

  Is there an error in this question or solution?

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Solution Find the Values of K for Which the System 2x + Ky = 1 3x – 5y = 7 Will Have (I) a Unique Solution, and (Ii) No Solution. is There a Value of K for Which the System Has Infinitely Many Solutions? Concept: Pair of Linear Equations in Two Variables.
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