Question

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

• A. 1
• B. 3
• C. 2
• D. infinite

Answer 1

The correct option is A 1

Let $△=\left|\begin{array}{cc}k+1& 8\\ k& k+3\end{array}\right|$

$=k^2+4k+3=(k-1)(k-3)$

For the system of equation to have no solution

$△=0\Rightarrow k=3,1$

For $k=1,$ the system of equation becomes

$2x+8y=4$ and $x+4y=2$

$\Rightarrow x+4y=2,x+4y=2$

This system has infinite number of solution

For $k=3$, the system of equation becomes

$4x+8y=12$ and $3x+6y=8$

$\Rightarrow x+2y=3,x+2y=\frac{8}{3}$

Thus, in this case the system of equation has no solution.

The number of values of k for which the given system of equations has no solution is 1.