It seems there might be a mistake in your equations as they appear to be incorrectly written. You have written:

1. \( 3xy - x + 3y = 93 \)
2. \( xy - x + 3y = 9 \)

If your intention is to solve a system of equations, please clarify if the equations should be as I interpreted them, or if there are any corrections needed.

Assuming those are the correct equations, let's solve them:

**Equation 1:**
\( 3xy - x + 3y = 93 \)

**Equation 2:**
\( xy - x + 3y = 9 \)

### Step 1: Solve Equation 2 for \( x \)
From Equation 2:
\[
xy - x + 3y = 9
\]
Rearranging gives:
\[
xy - x = 9 - 3y
\]
Factoring out \( x \):
\[
x(y - 1) = 9 - 3y
\]
So, we have:
\[
x = \frac{9 - 3y}{y - 1}
\]

### Step 2: Substitute \( x \) in Equation 1
Now substitute \( x \) into Equation 1:
\[
3\left(\frac{9 - 3y}{y - 1}\right)y - \frac{9 - 3y}{y - 1} + 3y = 93
\]

### Step 3: Simplify the equation
Let's simplify this step-by-step:
\[
3\left(\frac{9y - 3y^2}{y - 1}\right) - \frac{9 - 3y}{y - 1} + 3y = 93
\]
Combining the two fractions:
\[
\frac{27y - 9y^2 - (9 - 3y)}{y - 1} + 3y = 93
\]
\[
\frac{27y - 9y^2 - 9 + 3y}{y - 1} + 3y = 93
\]
\[
\frac{-9y^2 + 30y - 9}{y - 1} + 3y = 93
\]
Now multiply through by \( y - 1 \) to eliminate the denominator:
\[
-9y^2 + 30y - 9 + 3y(y - 1) = 93(y - 1)
\]
This becomes:
\[
-9y^2 + 30y - 9 + 3y^2 - 3y = 93y - 93
\]
Combine like terms:
\[
-6y^2 + 27y + 84 = 0
\]
Multiply by -1 to make it simpler:
\[
6y^2 - 27y - 84 = 0
\]

### Step 4: Solve the quadratic
You can either factor or use the quadratic formula. The quadratic formula says:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 6, b= -27, c=-84 \).

Calculate the discriminant:
\[
b^2 - 4ac = (-27)^2 - 4(6)(-84) = 729 + 2016 = 2745
\]

So,
\[
y = \frac{27 \pm \sqrt{2745}}{12}
\]

The square root of \( 2745 \) can be simplified to \( 3\sqrt{305} \):
\[
y = \frac{27 \pm 3\sqrt{305}}{12}
\]

This gives potential values for \( y \). After finding \( y \), substitute back to find \( x \) using the earlier derived formula.

If you are looking for specific numerical solutions, without the presence of radicals, please let me know!