{3х-1y=1 {2x+3y=16

To solve the system of equations:

1. \(3x - y = 1\) (Equation 1)
2. \(2x + 3y = 16\) (Equation 2)

First, we can solve Equation 1 for \(y\):

\[ y = 3x - 1 \]

Now, we can substitute this expression for \(y\) into Equation 2:

\[ 2x + 3(3x - 1) = 16 \]

Expanding this gives:

\[ 2x + 9x - 3 = 16 \]

Combining like terms:

\[ 11x - 3 = 16 \]

Now, add 3 to both sides:

\[ 11x = 19 \]

Dividing both sides by 11 gives:

\[ x = \frac{19}{11} \]

Now substitute \(x\) back into the equation we found for \(y\):

\[ y = 3\left(\frac{19}{11}\right) - 1 \]

Calculating that out:

\[ y = \frac{57}{11} - 1 = \frac{57}{11} - \frac{11}{11} = \frac{46}{11} \]

So the solution to the system of equations is:

\[ x = \frac{19}{11}, \quad y = \frac{46}{11} \]

This can also be expressed as the ordered pair:

\[ \left(\frac{19}{11}, \frac{46}{11}\right) \]