Solve the system by any method.
\[
\begin{array}{r}
x+3 y=5 \\
2 x+4 y=4 \\
3 x+5 y=3 \\
4 x+6 y=2
\end{array}
\]

Step 1 ：The given system of equations is: \[\begin{align*} 1) & x + 3y = 5 \\ 2) & 2x + 4y = 4 \\ 3) & 3x + 5y = 3 \\ 4) & 4x + 6y = 2 \end{align*}\]

Step 2 ：Simplify the equations. The second equation can be simplified by dividing the entire equation by 2: \[x + 2y = 2\] The fourth equation can be simplified by dividing the entire equation by 2: \[2x + 3y = 1\] Now, the system of equations is: \[\begin{align*} 1) & x + 3y = 5 \\ 2) & x + 2y = 2 \\ 3) & 3x + 5y = 3 \\ 4) & 2x + 3y = 1 \end{align*}\]

Step 3 ：Subtract the second equation from the first: \[(x + 3y) - (x + 2y) = 5 - 2\] which simplifies to \[y = 3\]

Step 4 ：Substitute \(y = 3\) into the first equation: \[x + 3*3 = 5\] which simplifies to \[x + 9 = 5\] and further simplifies to \[x = 5 - 9\] which gives \[x = -4\]

Step 5 ：So, the solution to the system of equations is \(x = -4\) and \(y = 3\)

Step 6 ：Check the solution by substituting \(x = -4\) and \(y = 3\) into the original equations: \[\begin{align*} 1) & -4 + 3*3 = 5, \text{ which is true.} \\ 2) & 2*(-4) + 4*3 = 4, \text{ which is true.} \\ 3) & 3*(-4) + 5*3 = 3, \text{ which is true.} \\ 4) & 4*(-4) + 6*3 = 2, \text{ which is true.} \end{align*}\]

Step 7 ：Therefore, the solution \(x = -4\) and \(y = 3\) is correct. The final answer is \(\boxed{x = -4, y = 3}\)