How do I know that this system of equations has infinitely many solutions? \begin{cases}
2x + 4y - 2z = 0\\
\\3x + 5y      = 1
\end{cases}
My book is using this as an example of a system of equations that has infinitely many solutions, but I want to know how we can know that just from looking at the equations?

One can write your system $A x = b$ as augmented matrix and bring it into row echelon form
$$
\left[
\begin{array}{rrr|r}
2 & 4 & -2 & 0 \\
3 & 5 &  0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 2 & -1 & 0 \\
3 & 5 &  0 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 &  2 & -1 & 0 \\
0 & -1 &  3 & 1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 2 & -1 &  0 \\
0 & 1 & -3 & -1
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 0 & 5 &  2 \\
0 & 1 & -3 & -1
\end{array}
\right]
$$
This translates back into
$$
x + 5z = 2 \\
y - 3z = -1
$$
or $x = (2-5z, -1+3z, z)^T$, where $z \in \mathbb{R}$.
So there are infinite many solutions.
From a geometric point of view, each equation defines an affine plane in $\mathbb{R}^3$, which is a plane, not necessarily including the origin.
$$
(2, 4, -2) \cdot (x,y,z) = 0 \\
(3, 5, 0) \cdot (x, y, z) = 1
$$
The first plane has a normal vector $(2,4,-2)^T$ and includes the origin,
the second plane has normal vector $(3,5,0)$ and is $1/\sqrt{3^2 + 5^2}$ away from the origin.

(Large version)
The solution of the system is the intersection of those two planes.
And only the cases empty intersection, intersection is a line or intersection is a plane can happen. Here the intersection is a line.
The image shows the two intersecting planes, the intersection line, and the point $P$ which corresponds to the solution with $z = 0$.
