Find all functions $ f,g:\mathbb{Q}\longrightarrow\mathbb{Q} $ that verify the relations
$$ \left\{\begin{matrix} f(g(x)+g(y))=f(g(x))+y \\
g(f(x)+f(y))=g(f(x))+y\end{matrix}\right. , $$
for all $ x,y\in\mathbb{Q} . $
To find all functions \( f, g: \mathbb{Q} \to \mathbb{Q} \) that satisfy the given relations:
\[
\begin{cases}
f(g(x) + g(y)) = f(g(x)) + y \\
g(f(x) + f(y)) = g(f(x)) + y
\end{cases}
\]
for all \( x, y \in \mathbb{Q} \), we proceed as follows:

1. **Show that \( f \) and \( g \) are bijective:**

   - **Injectivity of \( f \):**
     Suppose \( f(a) = f(b) \). Then, for any \( y \in \mathbb{Q} \),
     \[
     f(g(a) + g(y)) = f(g(a)) + y = f(g(b)) + y = f(g(b) + g(y)).
     \]
     Since \( f \) is a function, \( g(a) + g(y) = g(b) + g(y) \). By the cancellation law in \( \mathbb{Q} \), \( g(a) = g(b) \). Since \( g \) is also a function, \( a = b \). Hence, \( f \) is injective.

   - **Surjectivity of \( f \):**
     For any \( z \in \mathbb{Q} \), we need to find \( x \in \mathbb{Q} \) such that \( f(x) = z \). Since \( f \) is injective and \( \mathbb{Q} \) is countable, \( f \) must map \( \mathbb{Q} \) onto itself. Thus, \( f \) is surjective.

   - **Injectivity and Surjectivity of \( g \):**
     Similar arguments can be applied to show that \( g \) is both injective and surjective.

2. **Show that \( f(0) = g(0) = 0 \):**

   - Set \( x = 0 \) in the first equation:
     \[
     f(g(0) + g(y)) = f(g(0)) + y.
     \]
     Let \( c = f(g(0)) \). Then,
     \[
     f(g(0) + g(y)) = c + y.
     \]
     Since \( f \) is surjective, for any \( z \in \mathbb{Q} \), there exists \( y \in \mathbb{Q} \) such that \( g(y) = z - g(0) \). Thus,
     \[
     f(z) = c + y.
     \]
     This implies \( f \) is a linear function, and \( c = 0 \). Hence, \( f(g(0)) = 0 \), so \( g(0) = 0 \).

   - Set \( x = 0 \) in the second equation:
     \[
     g(f(0) + f(y)) = g(f(0)) + y.
     \]
     Let \( d = g(f(0)) \). Then,
     \[
     g(f(0) + f(y)) = d + y.
     \]
     Since \( g \) is surjective, for any \( z \in \mathbb{Q} \), there exists \( y \in \mathbb{Q} \) such that \( f(y) = z - f(0) \). Thus,
     \[
     g(z) = d + y.
     \]
     This implies \( g \) is a linear function, and \( d = 0 \). Hence, \( g(f(0)) = 0 \), so \( f(0) = 0 \).

3. **Show that \( f(g(x)) = g(f(x)) = x \):**

   - From the first equation, set \( y = 0 \):
     \[
     f(g(x) + g(0)) = f(g(x)) + 0 \implies f(g(x)) = f(g(x)).
     \]
     Since \( f \) is injective, \( g(x) = x \).

   - From the second equation, set \( y = 0 \):
     \[
     g(f(x) + f(0)) = g(f(x)) + 0 \implies g(f(x)) = g(f(x)).
     \]
     Since \( g \) is injective, \( f(x) = x \).

4. **Show that \( f \) and \( g \) are additive and linear:**

   - From the first equation:
     \[
     f(g(x) + g(y)) = f(g(x)) + y.
     \]
     Since \( g(x) = x \), this becomes:
     \[
     f(x + y) = f(x) + y.
     \]
     Let \( y = 0 \):
     \[
     f(x) = f(x) + 0.
     \]
     Thus, \( f \) is additive.

   - From the second equation:
     \[
     g(f(x) + f(y)) = g(f(x)) + y.
     \]
     Since \( f(x) = x \), this becomes:
     \[
     g(x + y) = g(x) + y.
     \]
     Let \( y = 0 \):
     \[
     g(x) = g(x) + 0.
     \]
     Thus, \( g \) is additive.

5. **Determine the linear forms of \( f \) and \( g \):**

   Since \( f \) and \( g \) are additive and map \( \mathbb{Q} \) to \( \mathbb{Q} \), they must be linear. Let \( f(x) = ax \) and \( g(x) = bx \). Substituting these into the original equations, we get:
   \[
   f(g(x) + g(y)) = f(bx + by) = a(bx + by) = ab(x + y),
   \]
   \[
   f(g(x)) + y = abx + y.
   \]
   Equating these, we get \( ab = 1 \).

   Similarly,
   \[
   g(f(x) + f(y)) = g(ax + ay) = b(ax + ay) = ba(x + y),
   \]
   \[
   g(f(x)) + y = bax + y.
   \]
   Equating these, we get \( ba = 1 \).

   Therefore, \( ab = 1 \) and \( ba = 1 \), implying \( a = b \) and \( a^2 = 1 \). Thus, \( a = 1 \) or \( a = -1 \).

   Hence, the solutions are:
   \[
   f(x) = x \quad \text{and} \quad g(x) = x,
   \]
   or
   \[
   f(x) = -x \quad \text{and} \quad g(x) = -x.
   \]

The final answer is \(\boxed{f(x) = ax \text{ and } g(x) = bx \quad \forall x}\) where \(a, b \in \mathbb{Q}\) such that \(ab = 1\).