Answer
If $h = 2$, then the system has no solution, otherwise, when $h \neq 2$, the system has a solution.
Work Step by Step
We begin with matrix:
$\begin{bmatrix}
1 & h & 4\\
3 & 6 & 8
\end{bmatrix}$
First, we need to multiply the first row by $-3$ and add it to the second row:
$\begin{bmatrix}
1 & h & 4\\
0 & 6-3h & -4
\end{bmatrix}$
Write $c$ for $6 – 3h$. If $c=0$ then $6-3h=0$. The solution of this equation is $h=2$.
If $c = 0$, that is, if $h = 2$, then the system has no solution, because $0$ cannot equal $–4$. Otherwise, when $h \neq 2$, the system has a solution.
