Here, the equations are;

$(k+2)x+10y=k$

and $ kx+(k+3)y=k-1$.

These equations can be written in the form of $Ax=Bas$

$\left[ {\begin{array}{*{20}{c}} {k + 2}&{10}\\ k&{k + 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} k\\ {k – 1} \end{array}} \right]$

For the system to have no solution

              $\left| A \right| = 0$

$ \Rightarrow \left[ {\begin{array}{*{20}{c}} {k + 2}&{10}\\ k&{k + 3} \end{array}} \right] = 0$

$ \Rightarrow \left( {k + 2} \right)\left( {k + 3} \right) – k \times 10 = 0$

$ \Rightarrow {k^2} – 5k + 6 = \left( {k – 2} \right)\left( {k – 3} \right) = 0$

$\therefore k = 2.3$

For $k=2,$ equations become:

$4x + 100y = 2$ and $2x + 5y = 1$

 hence infintie number of solutions.

For $k=3$, equations becomes;

$5x + 10y = 3$

$3x + 6y = 2$

$4$ hence no solution.

$\therefore $ req. number of values of $k$ is $1$.