Let's solve the given systems of equations one by one.

### System 50:
1. \( 4x + y = 18 \)
2. \( 2x + 3y = 4 \)

First, solve the first equation for \( y \):
\[ y = 18 - 4x \]

Substitute \( y \) in the second equation:
\[ 2x + 3(18 - 4x) = 4 \]
\[ 2x + 54 - 12x = 4 \]
\[ -10x + 54 = 4 \]
\[ -10x = 4 - 54 \]
\[ -10x = -50 \]
\[ x = 5 \]

Now, substitute \( x = 5 \) back into the first equation to find \( y \):
\[ 4(5) + y = 18 \]
\[ 20 + y = 18 \]
\[ y = 18 - 20 \]
\[ y = -2 \]

So, the solution to system 50 is:
\[ x = 5 \]
\[ y = -2 \]

### System 54:
1. \( 2x + y = 1 \)
2. \( 4x + 2y = 2 \)

First, solve the first equation for \( y \):
\[ y = 1 - 2x \]

Substitute \( y \) in the second equation:
\[ 4x + 2(1 - 2x) = 2 \]
\[ 4x + 2 - 4x = 2 \]
\[ 2 = 2 \]

This equation is always true, which means the system has infinitely many solutions. The second equation is just a multiple of the first equation. Therefore, the solutions are all points \((x, y)\) that satisfy the first equation:
\[ y = 1 - 2x \]

So, the solution to system 54 is:
\[ y = 1 - 2x \]
where \( x \) can be any real number.