Does this two equation system have a solution for l(1)? I'm having a debate with a friend over this problem:
$$\eqalign{
k(x) &= l(x)+8 \cr
l(x) &= 14-k(x)
}$$
What is the value of $l(1)$?
I think this isn't solvable because of the circular reference/dependency, but my friend thinks that it can be done with substitution, resulting in constant answers:
$$\eqalign{
l(x)&=14-[l(x)+8] \cr
2l(x)&=6 \cr
l(x)&=3
}$$
which leads to
$$\eqalign{
k(x)&=3+8 \cr
k(x)&=11
}$$
Thoughts?
 A: There is nothing circular in this problem since the two equations are different and cannot be derived from each other. Thus, using the substitution, it is true that $l(x)$ and $k(x)$ are always $3$ and $11$ respectively, which means that $l(1)$ is also $3$
A: 
I think this isn't solvable because of the circular reference/dependency

It's just a linear system of 2 equations in 2 variables, where it does not matter that $k$ and $\ell$ are functions.  One of the following cases will occur:


*The system has no solutions because it's over-determined.  Example would be if the equations are, say $\ell=1$ and $2\ell=-1$ etc.


*There is a unique solution


*There is a space of solutions, i.e. the solution is not unique.  The equations just limit the degree of freedom of the 2-dimensional space in which $k$ and $\ell$ live.
In your case, the system of equations is
$$\begin{align}
k - \ell &= 8 \\
k + \ell &= 14
\end{align}$$
so that adding these equations gives an equation for $k$, and subtracting the equations gives an equation for $\ell$.  It is case 1. from above, i.e. the solution is unique.
If $k$ and $\ell$ are functions (of $x$), then they only satisfy the system of equations if they are constant functions with the values you computed.
