Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying, for a given real $k$,

$$
\forall x, y \in \mathbb{R}, \quad f(2 x-2 y)+x=f(3 x)-f(2 y)+k y
$$

## Solutions of exercises
One of the main ideas in a functional equation with multiple terms is to make what is inside the parentheses equal, through a clever substitution, to obtain simplifications. For example, here, three substitutions are possible: $2 x - 2 y = 3 x$, $2 x - 2 y = 2 y$, or $3 x = 2 y$. We will only present the first one here, but it is worth noting that the third one also leads to a quick solution (try it!)

We want to have $2 x - 2 y = 3 x$. This is the case if we set $x = -2 y$. Let's see where this leads us:

$$
f(-6 y) - 2 y = f(-6 y) - f(2 y) + k y \Longleftrightarrow f(2 y) = (k + 2) y
$$

By setting $z = 2 y$, we get $f(z) = \frac{1}{2}(k + 2) z$

Let's test this solution:

$$
\begin{gathered}
\frac{1}{2}(k + 2)(2 x - 2 y) + x = \frac{1}{2}(k + 2) \times 3 x - \frac{1}{2}(k + 2) \times 2 y + k y \\
\Longleftrightarrow (2 k + 6) x - (2 k + 4) y = (3 k + 6) x - (k + 4) y \\
\quad \Longleftrightarrow (2 k + 6 = 3 k + 6 \text{ and } 2 k + 4 = k + 4) \Longleftrightarrow k = 0
\end{gathered}
$$

Therefore, there are solutions only for $k = 0$, in which case $\frac{1}{2}(k + 2) x = x$, and the function $f(x) = x$ is indeed a solution.

## Injectivity, Surjectivity, and Bijectivity

We now introduce the essential concepts of injective, surjective, and bijective functions. These are properties that we can try to deduce from the statement to get clues about the solutions.

## Definition 2.

A function is said to be injective if the images of two distinct elements by $f$ are distinct. In other words, if $a$ and $b$ belong to the domain of definition of $f$ and $a \neq b$, then $f(a) \neq f(b)$. Thus, every element in the codomain has a unique preimage.

## Definition 3.

A function is said to be surjective if every element in the codomain has at least one preimage. In other words, if $f: A \rightarrow B$, then for every $y \in B$ there exists an $x \in A$ such that $f(x) = y$.

## Definition 4.

A function is said to be bijective if it is both injective and surjective.

## Remark 5.

A function is therefore bijective if every element in the codomain has a unique preimage.

## Remark 6.

To show that a function is injective, we often use the contrapositive: we show that if two elements $a$ and $b$ satisfy $f(a) = f(b)$, then $a = b$.

## Remark 7.

To show that a function is surjective, we generally try to reduce it to an equation of the form $f(A) = B$, where $A$ and $B$ are expressions that depend on a variable $x$. If $B$ covers the entire codomain of $f$ as $x$ covers the same set, then $f$ is surjective.