{3x-5y=2 -x+4y=4

To solve the system of equations:

1. \(3x - 5y = 2\) (Equation 1)
2. \(-x + 4y = 4\) (Equation 2)

We can use the substitution or elimination method. Here, I will use the elimination method.

First, we can multiply Equation 2 by 3 to align the \(x\) coefficients:

\[ 3(-x + 4y) = 3(4) \implies -3x + 12y = 12 \quad (Equation 3) \]

Now we have:

1. \(3x - 5y = 2\) (Equation 1)
2. \(-3x + 12y = 12\) (Equation 3)

Now we can add Equation 1 and Equation 3:

\[ (3x - 5y) + (-3x + 12y) = 2 + 12 \]

This simplifies to:

\[ (3x - 3x) + (-5y + 12y) = 14 \]

\[ 0 + 7y = 14 \]

Now, isolate \(y\):

\[ 7y = 14 \implies y = 2 \]

Now that we have \(y\), we can substitute it back into one of the original equations to find \(x\). We'll use Equation 2:

\[ -x + 4(2) = 4 \]

This simplifies to:

\[ -x + 8 = 4 \]

Now, isolate \(x\):

\[ -x = 4 - 8 \]

\[ -x = -4 \implies x = 4 \]

Thus, the solution to the system of equations is:

\[ \boxed{(4, 2)} \]