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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)

best_answer

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.

Posted by

Aadil

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