System of three equations 
$$3x+2y-z=4$$ $$4y-2z+6x=8$$ $$x-2y=5$$

I found that the first two equations were the same. Then, looking at the second and third equations, I thought the answer was No Solution, because I couldn't find a way to solve it, but instead the answer was Infinitely Many Solutions. Can someone explain this to me?
 A: The system depends on the arbitrary choice of one of the unknowns. Indeed, it is equivalent to 
$$\begin{cases}3x-2y-z=4\\x-2y=5\end{cases}\iff\begin{cases} x=5+2y\\z=-4+3x+2y=11+8y\end{cases}$$
So the set of solutions is
$$\begin{pmatrix}
x\\y\\z\end{pmatrix}=\begin{pmatrix}
5\\0\\11\end{pmatrix}+y\begin{pmatrix}
2\\1\\8\end{pmatrix}.$$
A: This is the same as$$\begin{bmatrix}3&2&-1\\6&4&-2\\4&-2&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\8\\5\end{bmatrix}$$
Since $\begin{bmatrix}3&2&-1\\6&4&-2\\4&-2&0\end{bmatrix}$ is non-invertable, it suggests 0 or infinite solutions.
Now we just have to find 1 solution to show it has infinite solutions.
Letting $y=0$, from $x-2y=5$, $x=5$, and then from $3x+2y-z=5$, $z=11$.
Generating solutions:
We can clearly see the first and second equations are identical, so we may forget about the second. 
$$x=5+2y$$
Now plugging this into the first equation
$$z=3(5+2y)+2y-5=8y+10$$
A: As one of the equations is a duplicate, you can drop it and solve.
$$\begin{cases}3x+2y-z=4\\
x-2y=5\end{cases}$$
Let us move $z$ to the RHS and consider it as known to obtain a system of two equations in two unknowns,
$$\begin{cases}3x+2y=z+4\\
x-2y=5.\end{cases}$$
Then by elimination
$$\begin{cases}4x=z+9\\
8y=z-11.\end{cases}$$
We can assign $z$ any value, yet get values of $x,y,z$ that satisfy all equations. For instance $(3,-1,3)$ or $(1,-2,-5)$. There are infinitely many others. In the solution space, $\mathbb R^3$, the solutions describe a straight line.
A: Let $z=t$ and removing the second equations, your problem reduced to 
$$3x+2y=4+t$$
$$x-2y=5$$
Summing the two equations up, we have
$$4x=9+t$$
and we can conclude that $$x=\frac{9+t}{4}$$
Can you use $x=\frac{9+t}{4}$ and $x-2y=5$ to express $y$ in terms of $t$?
Whenever you fix a value of $t$, you obtain a different solutions of $(x,y,z)$, hence there are infinitely many solutions.
