Find all pairs $(x, y)$ of real numbers satisfying the system :
$\begin{cases} x + y = 2 \\
x^4 - y^4 = 5x - 3y \end{cases}$
1. Start with the given system of equations:
   \[
   \begin{cases}
   x + y = 2 \\
   x^4 - y^4 = 5x - 3y
   \end{cases}
   \]

2. From the first equation, solve for \( y \):
   \[
   x + y = 2 \implies y = 2 - x
   \]

3. Substitute \( y = 2 - x \) into the second equation:
   \[
   x^4 - (2 - x)^4 = 5x - 3(2 - x)
   \]

4. Expand \( (2 - x)^4 \):
   \[
   (2 - x)^4 = 16 - 32x + 24x^2 - 8x^3 + x^4
   \]

5. Substitute the expanded form back into the equation:
   \[
   x^4 - (16 - 32x + 24x^2 - 8x^3 + x^4) = 5x - 3(2 - x)
   \]

6. Simplify the left-hand side:
   \[
   x^4 - 16 + 32x - 24x^2 + 8x^3 - x^4 = 5x - 6 + 3x
   \]
   \[
   8x^3 - 24x^2 + 32x - 16 = 8x - 6
   \]

7. Move all terms to one side of the equation:
   \[
   8x^3 - 24x^2 + 32x - 16 - 8x + 6 = 0
   \]
   \[
   8x^3 - 24x^2 + 24x - 10 = 0
   \]

8. Factor the polynomial:
   \[
   8(x^3 - 3x^2 + 3x - \frac{5}{4}) = 0
   \]

9. Solve the cubic equation:
   \[
   x^3 - 3x^2 + 3x - \frac{5}{4} = 0
   \]

10. Use the Rational Root Theorem or numerical methods to find the roots. One root is:
    \[
    x = 3 + \frac{1}{\sqrt[3]{4}}
    \]

11. Substitute \( x = 3 + \frac{1}{\sqrt[3]{4}} \) back into \( y = 2 - x \):
    \[
    y = 2 - \left(3 + \frac{1}{\sqrt[3]{4}}\right) = -1 - \frac{1}{\sqrt[3]{4}}
    \]

12. Verify the solution by substituting \( x \) and \( y \) back into the original equations to ensure they are satisfied.

The final answer is \( \boxed{ \left(3 + \frac{1}{\sqrt[3]{4}}, -1 - \frac{1}{\sqrt[3]{4}}\right) } \)