Let's solve the systems of equations given in the image.

### System 51:
\[
\begin{cases}
x + 2y = 5 \\
3x + 4y = 5
\end{cases}
\]

1. Multiply the first equation by 2 to align the coefficients of \(y\):
\[
2(x + 2y) = 2 \cdot 5 \implies 2x + 4y = 10
\]

2. Now we have:
\[
\begin{cases}
2x + 4y = 10 \\
3x + 4y = 5
\end{cases}
\]

3. Subtract the second equation from the first:
\[
(2x + 4y) - (3x + 4y) = 10 - 5 \implies -x = 5 \implies x = -5
\]

4. Substitute \(x = -5\) back into the first equation:
\[
-5 + 2y = 5 \implies 2y = 10 \implies y = 5
\]

So, the solution for system 51 is \(x = -5\) and \(y = 5\).

### System 55:
\[
\begin{cases}
x - 2y = 4 \\
2x - 4y = 1
\end{cases}
\]

1. Notice that the second equation is a multiple of the first equation:
\[
2(x - 2y) = 2 \cdot 4 \implies 2x - 4y = 8
\]

2. Now we have:
\[
\begin{cases}
x - 2y = 4 \\
2x - 4y = 1
\end{cases}
\]

3. Subtract the second equation from the modified first equation:
\[
(2x - 4y) - (2x - 4y) = 8 - 1 \implies 0 = 7
\]

This is a contradiction, which means that the system of equations is inconsistent and has no solution.

So, the solution for system 55 is that there is no solution.