The number of values of k, for which the system of equations :

(k+1)x+8y=4k

kx+(k+3)y=3k-1

has no solution, is :

 

 

  • Option 1)

    3

  • Option 2)

    infinite

  • Option 3)

    1

  • Option 4)

    2

 

Answers (2)

As we learnt in 

Cramer's rule for solving system of linear equations -

When \Delta =0 and atleast one of   \Delta_{1},\Delta _{2} and \Delta _{3}  is non-zero , system of equations has no solution

- wherein

a_{1}x+b_{1}y+c_{1}z=d_{1}

a_{2}x+b_{2}y+c_{2}z=d_{2}

a_{3}x+b_{3}y+c_{3}z=d_{3}

and 

\Delta =\begin{vmatrix} a_{1} &b_{1} &c_{1} \\ a_{2} & b_{2} &c_{2} \\ a_{3}&b _{3} & c_{3} \end{vmatrix}

 

\left ( k+1 \right )x+8y=4k

kx + \left ( k+3 \right )y = 3k-1

\frac{k+1}{k} = \frac{8}{k+3} \neq \frac{4k}{3k-1}

\therefore \left ( k+1 \right )\left ( k+3 \right )= 8k

\therefore k^{2}+4k+3-8k=0

\therefore k^{2}-4k+3=0

\Rightarrow \left ( k-1 \right )\left ( k-3 \right )=0

\Rightarrow k= 1\ and\ 3

But K=1 not satisfied

\therefore k=3 

 


Option 1)

3

This option is correct.

Option 2)

infinite

This option is incorrect.

Option 3)

1

This option is incorrect.

Option 4)

2

This option is incorrect.

N neha

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