Finding a solution to a system Let $i$ be of the form $i=2^a3^b5^c$, where $a,b,c\ge 0$ are integers. 
Consider numbers $x_{i,4},x_{i,6}$ where $x_{i,6}$ is defined when $2$ or $3$ divides $i$, $x_{i,4}$ is defined only when $2$ divides $i$.
The constraints are 


*

*If $2$ divides $i$, then $$x_{i, 4} + x_{i,6} \le \frac{\frac{1}{2}\frac{2}{3}\frac{4}{5}}{i}$$

*If $3$ divides $i$ but $2$ doesn't divide $i$, then $$x_{i,6} \le \frac{\frac{1}{2}\frac{2}{3}\frac{4}{5}}{i}$$

*$$\sum x_{i,4} = \sum x_{i,6} = \frac{1}{6}.$$


I know that this system has a solution, but the technique doesn't work when I include more $x_{i,j}$'s. I've never solved systems of this type; what's a good way to approach this problem?

Actually this system have no solution, first $x_2,y_2,x_3,y_3,x_5,y_5 \in \mathbb{R} \geq 0$.
And $x_2+y_2=1 = \sum \limits_{k=1}^{\infty} \frac{1}{2^k}$
And $x_3+y_3 = \frac{1}{2} =  \sum \limits_{k=1}^{\infty} \frac{1}{3^k}$
And $x_5 +y_5 = \frac{1}{4} = \sum \limits_{k=1}^{\infty} \frac{1}{5^k}$.
So $\sum x_{i,4} = \frac{1}{2}\frac{2}{3}\frac{4}{5}* \frac{1}{2} * ((x_2+1)(x_3+1)(x_5+1)-(x_3+1)(x_5+1))$ because if $2\not|i$ then its not defined for $x_{i,4}$.
And $\sum x_{i,6} = \frac{1}{2}\frac{2}{3}\frac{4}{5}* \frac{1}{3} * ((y_2+1)(y_3+1)(y_5+1)-(y_5+1))$ because if $2\not|i$ and $3\not|i$ then its not defined for $x_{i,6}$.
Solving this system of equations using Wolfram|Alpha (also simplex method works) yield that there is no solution for your system.
