System of Equations with Two variables I am stuck on the following question:
Consider the system of equations
$$2x + by = 1$$
$$3x + y  = c$$
For which values of b and c does the system have a unique solution?
For which values of b and c is the system inconsistent?
The internet tells me that the determinant cannot equal $0$ but I'm not sure how to apply this with the two constants
 A: The coefficient matrix is $\begin{pmatrix}2 & b \\ 3 & 1 \end{pmatrix}$, so the determinant is $2-3b$. The determinant is nonzero when $b\neq \frac{2}{3}$. Then $c$ can be anything and there will be a unique solution.
As for your second question: when $b=\frac{2}{3}$, you can perform row reduction to find out that the system is inconsistent when $c \neq \frac{3}{2}$.
Conclusion:
For which values of b and c does the system have a unique solution? Any as long as $b\neq \frac{2}{3}$.
For which values of b and c is the system inconsistent? Inconsistent when $b=\frac{2}{3}$ and $c \neq \frac{3}{2}$.
A: You can re-write these equations as a matrix equation:
$$\left(\begin{array}{cc} 2 & b \\ 3 & 1 \end{array}\right)\left(\begin{array}{c} x \\ y \end{array}\right)=\left(\begin{array}{c} 1 \\ c \end{array}\right)$$
If the two-by-two matrix on the left is non-singular, i.e. has non-zero determinant, then we can multiply on the left by the inverse matrix: 
$$M{\bf x} = {\bf a} \implies M^{-1}M{\bf x}=M^{-1}{\bf a} \implies I{\bf x} = M^{-1}{\bf a} \implies {\bf x} = M^{-1}{\bf a}$$
In our case, the determinant is $2-3b$. If $2-3b \neq 0$ then the inverse matrix exists and we can find a unique solution for ${\bf x}$.
$$\left(\begin{array}{c} x \\ y \end{array}\right)=\frac{1}{2-3b}\left(\begin{array}{cc} 1 & -b \\ -3 & 2 \end{array}\right)\left(\begin{array}{c} 1 \\ c \end{array}\right)$$
$$x=\frac{1-bc}{2-3b} \ \ \ \ \ \ \ \ y = \frac{2c-3}{2-3b}$$
A: Make the system of equation is a matrix-vector equation
$$ \begin{vmatrix} 2 & b \\ 3 & 1 \end{vmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ c \end{bmatrix} $$
Now you can see under what conditions you have two identical equations and when you have two parallel lines (each equation being a line in the xy plane).
Edit 1
If the determinant is zero $2-3b=0$, $b=\frac{2}{3}$ the two equations have equivalent left hand size
$$\left. \begin{aligned} 2 x + \frac{2}{3} y & =1 \\ 3 x + y & = c \end{aligned} \right\} \left. \begin{aligned} \frac{3}{2} \left( 2 x + \frac{2}{3} y \right) & =\frac{3}{2} (1) \\ 3 x + y & = c \end{aligned} \right\} \begin{aligned} 3 x +  y & = \frac{3}{2} \\ 3 x + y & = c \end{aligned} $$
So your two cases are $c=\frac{3}{2}$ and $c \neq \frac{3}{2}$. 
The first makes both equations the same with infinite solutions. For example $$\begin{aligned} x & = \frac{t}{3} \\ y & = \frac{1}{2}-t \end{aligned}$$
The second makes the equations inconsistent. 
When $b \neq \frac{2}{3}$ all values of $c$ produce a result.
