What $h$ and $k$ would make this system have infinitely many solutions? If there are an infinite number of solutions to the system 
$$\left|\begin{array}{cc|c} -5 & 6 & h\\ -8 & k & -7\end{array}\right|$$
then what do $h$ and $k$ equal?

I know that for a system to have infinity many solutions then both of the rows must be equal.
 A: I assume the system you have is
$$\begin{bmatrix} -5 & 6\\ -8 & k\end{bmatrix} \begin{bmatrix}x_1\\ x_2 \end{bmatrix} = \begin{bmatrix} h \\ -7\end{bmatrix}$$
For this to have infinite solutions, the two rows must be linearly dependent i.e. the second row must be a scalar multiple of the first row.
Note that $- 8 = - 5 \times \dfrac85$. Hence, $k = 6 \times \dfrac85 = \dfrac{48}5$ and $-7 = \dfrac85 \times h \implies h = - \dfrac{35}8$.
A: the rows of matrix are proportional
$$(-5):(-8)=6:k=h:(-7)$$
$$-5k=-48,h=35/-8$$
$$k=\frac{48}{5},h=-\frac{35}{8}$$
A: If you have the system of two lines: $$\begin{cases}
  ax+by=c \\
  a'x+b'y=c'
\end{cases}$$ then the system is infinite solutions iff we have: $$\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}$$ Of course we have some restrictions on the coefficients.
A: As you say, the rows must be equal. They should give you the same information, which occurs only if one is a scalar multiple of the other. Therefore
$\frac{-5}{-8}=\frac{6}{k}=\frac{h}{-7}$. This gives you $k=\frac{48}{5}$ and $h=\frac{-35}{8}$. 
