Here is a riddle that I have no idea how to solve. Okay, so I was trying to solve this riddle found here. 
It is a diagram of a star with 16 points. Each point corresponds uniquely to a number between 1 and 16. The letters on each point represent a letter of some saying, where if the unique corresponding to a point is $n$, then the letter on that point is the $n$th letter of the saying. We are also given the condition that the sum of the 4 numbers on a given line segment is the same.
I noticed that $\sum^{16}_{i=1} i = 136$, and each point is counted exactly twice, so each line should add up to $2\cdot136/8 = 34$. So we can find 8 equations in this way.

$$x_1 + x_2 + x_3 + x_4 = 34 $$
$$x_4 + x_5 + x_6 + x_7 = 34 $$
$$x_2 + x_7 + x_8 + x_9 = 34 $$
$$x_5 + x_9 + x_{10} + x_{11} = 34 $$
$$x_8 + x_{11} + x_{12} + x_{13} = 34 $$
$$x_{10} + x_{13} + x_{14} + x_{15} = 34 $$
$$x_3 + x_{12} + x_{15} + x_{16} = 34 $$
$$x_1 + x_6 + x_{14} + x_{16} = 34 $$
So, I have 8 equations and 16 unknowns, and finding the coefficient matrix and putting it in rref didn't shed that much light on the matter, because there are still too many unknowns. 
Now, I know there will be more than one solution. We can rotate it 7 times and there are 8 lines of symmetry; however, I think we should be able to narrow down our solutions farther than what I have already done. 
Does anyone have other ideas on how to approach this problem?
 A: Okay, the solution to this riddle is actually written in the blog you found it, but since you are asking about ideas on how to approach, i'll give u mine.
I found this solution (using GLPK)
$$
\begin{array}{cccc}
x_1 = 13 & x_2 = 8 & x_3 = 7 & x_4 = 6 \\
x_5 = 12 & x_6 = 2 & x_7 = 14 & x_8 = 11 \\
x_9 = 1 & x_{10} = 5 & x_{11} = 16 & x_{12} = 3 \\
x_{13} = 4 & x_{14} = 10 & x_{15} = 15 & x_{16} = 9
\end{array}
$$
where im using the same notation you did above. I also tried to figure out what was that saying, but i get a nosense group of letters, which I guess its because there exist, as u noticed, at least 8 simetric solutions. I saw in the blog some people solved the riddle just looking at the letters, but I am more of numbers, so I'll just skip that "saying" part.
Linear algebra part
My aproach consists on using the 8 conditions you already wrote, but adding the condition that each number between 1 and 16 must appear exactly once. Also we have to face the problem that our variables are integers, not real numbers. I dont even know if its possible to get such conditions using only linear algebra equations, a start would be adding this
$$
\sum_{i=1}^{16} x_i = 136
$$
but we still lack equations and i dont know how to proceed.
So I have used linear programming, and GLPK to solve it given this conditions.
Linear programming and aproach used (a bit long)
I dont know if you are familiar with linear programing, but I will explain the procedure in a way that it can be understood, hopefully.
Basicaly we define a new set of variables $x_{ij}$ which are binary (i.e $0$ or $1$) and are related to your variables this way
$$
x_i = \sum_{j=1}^{16} x_{ij}\cdot j
$$
Of course, then we only allow one of this $x_{ij}$ to be diferent from zero for each $i$, becasue "every unknown is only one number" so we add the condition
$$
\sum_{j=1}^{16} x_{ij} = 1
$$
The same way, we also want "each number to appear only once", which gives a very similar condition (note the change on the index of the sum)
$$
\sum_{i=1}^{16} x_{ij} = 1
$$
Finally, we substitute this new $x_{ij}$ on the 8 equations we had at the start and run GLPK. It tries to find a feasible solution, and it succeeds, giving the one i wrote above. I guess on diferent machines the program might find diferent solutions, since we already know its not unique.
Extra (I almost found the saying)
I was a bit dissapointed i didnt get the saying, so i decided to cheat a bit, took a look on the blog, the text is "do a good turn daily". His solution involves Y being the last letter, which means $x_{12}=16$. So i put this restraint on my program and... still got a diferent solution.
$$
\begin{array}{cccc}
x_1 = 8 & x_2 = 4 & x_3 = 15 & x_4 = 7 \\
x_5 = 6 & x_6 = 12 & x_7 = 9 & x_8 = 10 \\
x_9 = 11 & x_{10} = 14 & x_{11} = 3 & x_{12} = 16 \\
x_{13} = 5 & x_{14} = 13 & x_{15} = 2 & x_{16} = 1
\end{array}
$$
Not sure how many solutions this annoying star has :)
