a) $$ A=\{2 ; x ; 8\} $$ e $$ B=\{2 ; 5 ; y\} $$ b) $$ A=\{-1 ; 2 x+3 ; 7\} $$ e $$ B=\{y-4 ; 9 ; 7\} $$ c) $$ A=\{3 x-7 ;-5 ; 8 y-1\} $$ e $$ B=\{11 ;-5 ; 10 y-9\} $$ d) $$ A=\{x-y ; 5 ; 9\} $$ e $$ B=\{3 ; 3 x+y ; 9\} $$ e) $$ A=\{-3,3 x-4 y ; 1\} $$ e $$ B=\{-3 ; 2 ; 5 x-2 y\} $$

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To solve the problems, we need to find the values of the variables $$x$$ and $$y$$ in each pair of sets $$A$$ and $$B$$ such that the sets are equal. This means that every element in set $$A$$ must be equal to some element in set $$B$$.

Let's analyze each case step by step.

### a) $$ A=\{2 ; x ; 8\} $$ e $$ B=\{2 ; 5 ; y\} $$

1. From the sets, we can see that $$2$$ is common in both sets.
2. For the sets to be equal, the remaining elements must also match:
   - $$x$$ must equal $$5$$ (from set $$B$$).
   - $$8$$ must equal $$y$$ (from set $$B$$).
   
Thus, we have:
- $$x = 5$$
- $$y = 8$$

### b) $$ A=\{-1 ; 2x+3 ; 7\} $$ e $$ B=\{y-4 ; 9 ; 7\} $$

1. The common element is $$7$$.
2. The remaining elements must match:
   - $$-1$$ must equal $$y - 4$$ (from set $$B$$).
   - $$2x + 3$$ must equal $$9$$ (from set $$B$$).
   
Now we can set up the equations:
1. $$y - 4 = -1$$ → $$y = 3$$
2. $$2x + 3 = 9$$ → $$2x = 6$$ → $$x = 3$$

Thus, we have:
- $$x = 3$$
- $$y = 3$$

### c) $$ A=\{3x-7 ; -5 ; 8y-1\} $$ e $$ B=\{11 ; -5 ; 10y-9\} $$

1. The common element is $$-5$$.
2. The remaining elements must match:
   - $$3x - 7 = 11$$ (from set $$B$$).
   - $$8y - 1 = 10y - 9$$ (from set $$B$$).
   
Now we can set up the equations:
1. $$3x - 7 = 11$$ → $$3x = 18$$ → $$x = 6$$
2. $$8y - 1 = 10y - 9$$ → $$2y = 8$$ → $$y = 4$$

Thus, we have:
- $$x = 6$$
- $$y = 4$$

### d) $$ A=\{x-y ; 5 ; 9\} $$ e $$ B=\{3 ; 3x+y ; 9\} $$

1. The common element is $$9$$.
2. The remaining elements must match:
   - $$x - y = 3$$ (from set $$B$$).
   - $$5 = 3x + y$$ (from set $$B$$).
   
Now we can set up the equations:
1. $$x - y = 3$$ → $$y = x - 3$$
2. Substituting $$y$$ into the second equation: 
   $$5 = 3x + (x - 3)$$ → $$5 = 4x - 3$$ → $$4x = 8$$ → $$x = 2$$
3. Now substituting $$x$$ back to find $$y$$:
   $$y = 2 - 3 = -1$$

Thus, we have:
- $$x = 2$$
- $$y = -1$$

### e) $$ A=\{-3, 3x-4y ; 1\} $$ e $$ B=\{-3 ; 2 ; 5x-2y\} $$

1. The common element is $$-3$$.
2. The remaining elements must match:
   - $$3x - 4y = 2$$ (from set $$B$$).
   - $$1 = 5x - 2y$$ (from set $$B$$).
   
Now we can set up the equations:
1. $$3x - 4y = 2$$ (Equation 1)
2. $$5x - 2y = 1$$ (Equation 2)

We can solve these equations simultaneously.

Let's solve the system of equations:

$$
\begin{align*}
3x - 4y &= 2 \quad (1) \
5x - 2y &= 1 \quad (2)
\end{align*}
$$

We can multiply Equation (1) by $$2$$ and Equation (2) by $$4$$ to eliminate $$y$$:

$$
\begin{align*}
6x - 8y &= 4 \quad (3) \
20x - 8y &= 4 \quad (4)
\end{align*}
$$

Now, subtract Equation (3) from Equation (4):

$$
20x - 6x = 4 - 4 \
14x = 0 \
x = 0
$$

Now substituting $$x = 0$$ back into Equation (1):

$$
3(0) - 4y = 2 \
-4y = 2 \
y = -\frac{1}{2}
$$

Thus, we have:
- $$x = 0$$
- $$y = -\frac{1}{2}$$

### Summary of Results:
- a) $$x = 5$$, $$y = 8$$
- b) $$x = 3$$, $$y = 3$$
- c) $$x = 6$$, $$y = 4$$
- d) $$x = 2$$, $$y = -1$$
- e) $$x = 0$$, $$y = -\frac{1}{2}$$
- a) $$x = 5$$, $$y = 8$$ - b) $$x = 3$$, $$y = 3$$ - c) $$x = 6$$, $$y = 4$$ - d) $$x = 2$$, $$y = -1$$ - e) $$x = 0$$, $$y = -\frac{1}{2}$$

a) \( A=\{2 ; x ; 8\} \) e \( B=\{2 ; 5 ; y\} \) b) \( A=\{-1 ; 2 x+3 ; 7\} \) e \( B=\{y-4 ; 9 ; 7\} \) c) \( A=\{3 x-7 ;-5 ; 8 y-1\} \) e \( B=\{11 ;-5 ; 10 y-9\} \) d) \( A=\{x-y ; 5 ; 9\} \) e \( B=\{3 ; 3 x+y ; 9\} \) e) \( A=\{-3,3 x-4 y ; 1\} \) e \( B=\{-3 ; 2 ; 5 x-2 y\} \)

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