` (x + y - 8) /( 2 ) = ( x + 2y - 14 ) /( 3) = ( 3x + y - 12)/( 11)`

Explanation

To solve the given problem, we have the following equations:

\[ \frac{x + y - 8}{2} = \frac{x + 2y - 14}{3} = \frac{3x + y - 12}{11} \]

We will solve this by breaking it down into two parts.

### Step 1: Set up the first equation
From the first two fractions, we can set up the equation:

\[ \frac{x + y - 8}{2} = \frac{x + 2y - 14}{3} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:

\[ 3(x + y - 8) = 2(x + 2y - 14) \]

### Step 3: Expand both sides
Expanding both sides:

\[ 3x + 3y - 24 = 2x + 4y - 28 \]

### Step 4: Rearrange the equation
Rearranging the equation to group like terms:

\[ 3x - 2x + 3y - 4y = -28 + 24 \]

This simplifies to:

\[ x - y + 4 = 0 \]

### Step 5: Set up the second equation
Now, we will set up the second equation from the last two fractions:

\[ \frac{x + 2y - 14}{3} = \frac{3x + y - 12}{11} \]

### Step 6: Cross-multiply again
Cross-multiplying gives us:

\[ 11(x + 2y - 14) = 3(3x + y - 12) \]

### Step 7: Expand both sides
Expanding both sides:

\[ 11x + 22y - 154 = 9x + 3y - 36 \]

### Step 8: Rearrange the equation
Rearranging the equation:

\[ 11x - 9x + 22y - 3y = -36 + 154 \]

This simplifies to:

\[ 2x + 19y - 118 = 0 \]

### Step 9: Solve the system of equations
Now we have the system of equations:

1. \( x - y + 4 = 0 \) (Equation 1)
2. \( 2x + 19y - 118 = 0 \) (Equation 2)

From Equation 1, we can express \( x \) in terms of \( y \):

\[ x = y - 4 \]

### Step 10: Substitute into Equation 2
Substituting \( x \) into Equation 2:

\[ 2(y - 4) + 19y - 118 = 0 \]

### Step 11: Simplify and solve for \( y \)
Expanding gives:

\[ 2y - 8 + 19y - 118 = 0 \]

Combining like terms:

\[ 21y - 126 = 0 \]

Solving for \( y \):

\[ 21y = 126 \implies y = \frac{126}{21} = 6 \]

### Step 12: Substitute back to find \( x \)
Now substituting \( y \) back into the expression for \( x \):

\[ x = 6 - 4 = 2 \]

### Final Solution
Thus, the solution to the equations is:

\[ x = 2, \quad y = 6 \]

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