### JEE Main On 16 April 2018 Question 14

##### Question: The number of values of k for which the system of linear equations, $ (k+2)x+10y=k,kx+(k+3)y=k-1 $ has no solution, is? [JEE Main 16-4-2018]

#### Options:

A) Infinitely many

B) 3

C) 1

D) 2

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

A non-homogeneous system of linear equations has a unique non-trivial solution if and only if its determinant is non-zero.

If this determinant is zero, then the system has either no nontrivial solutions or an infinite number of solutions.

$ \begin{bmatrix} (k+2) & 10 \\ k & (k+3) \\ \end{bmatrix} $ $ [ \frac{x}{y} ]=[ \frac{k}{k-1} ] $

Now it is of the Form $ Ax=B $ Now to for the system to have no Solution, determinant of A must be 0, as follows

$ \Rightarrow $ $ |A|=(k+2)(k+3)-k\times 10=0\Rightarrow k^{2}-5k+6= $ $ (k-2)(k-3)=0 $

Therefore for k = 2, 3 system will have no solution. For k = 2,

we get infinitely many solutions, after substituting the value of k = 2 in the equations.

Thus $ k=3 $ Thus, the number of solutions is 1.