Simple system dilemma

Of course, if you graph both equations you can conclude that the system has a unique solution for whatever value of $k,$ but we cannot rely just on graphs an intuition. That is why we have to prove two things: the existence and the uniqueness of the solution.

Existence: It is easy because substituting the $y$ in the first equation by $-x+k$ and multiplying both sides by $1+x^2$ we obtain a real coefficient 3rd degree polynomial equation that always has at least a real zero. Let us say that the real zero is $a$ , then considering $b=-a+k$ , the pair $(a, b)$ will be a possible solution.

Uniqueness: First we can notice that the derivative of $y_1 =\frac{3x}{1+x^2,}$ which is $y_1'=-3\frac{x^2-1}{(x^2+1)^2},$ is always different from -1. Indeed, if you make $-3\frac{x^2-1}{(1+x^2)^2}=-1,$ this equation would be equivalent to the bi-quadratic equation $x^4-x^2+4=0$ that does not have real solutions.

Now, let us assume that the system has two different solutions: $(a_1, b_1)$ and $(a_2, b_2).$ Of course these two points are going to be on the line $y_2=-x+k$ , so $\frac{b_2-b_2}{a_2-a_1}=-1. \:\:\:\:\;\:\:\:\;\:\:(*)$ Since the two points are going to be also on the graph of $y_1 =\frac{3x}{1+x^2},$ then $(*)$ can be written in the following way: $\frac{y_1(a_2)-y_1(a_1)}{a_2-a_1}=-1.$ Using the Mean Value Theorem we would get the existence of number $c$ in between $a_1$ and $a_2$ such that $y_1'(c)=-1$ and this contradicts what we proved above. Therefore the system can not have more than one solution.

Then, we have proved that the system has unique solution for any real value of $k.$