> 
> Find the number of values of $k$ for which the system of equations has no solution:
>  $$(k+1)x+8y=4k$$
>  $$kx+(k+3)y=3k-1$$
> 
> 
> 

This is the augmented matrix:
$$\begin{bmatrix} 
k+1 & 8 & 4k \\
k & k+3 & 3k-1
\end{bmatrix}$$

$R\_2 \rightarrow R\_2-\frac{k}{k+1}R\_1$

$$\begin{bmatrix} 
k+1 & 8 & 4k \\
0 & k+3-\frac{8k}{k+1} & 3k-1-\frac{4k^2}{k+1}
\end{bmatrix}$$

If in the last row, the coefficient of $y$ where to become zero, keeping the constant term non zero, then there will be no solution. That is,
$$k+3-\frac{8k}{k+1}=0\text{ and } 3k-1-\frac{4k^2}{k+1}\neq 0$$

The only solution is $3$. But the answer says there are 3 solution for $k$.