Consider an farm where only wheat and cow (meat) is produced. For the sake of simplicity we will measure the amount of wheat and cow (meat) produced in tons (T). In order to produce w T of wheat one requires w/4 tons of wheat (to be used as seed), w/10 T of cow (which yields the required amount of cow manure to be used as fertilizer) and w/5 hrs of labor. Production c T of cow requires 2c T of wheat (to feed the animals), c/2 T of cow (to breed), and 4c hrs of labor. For workers to provide l hrs of labor the require l/4 T of wheat and l/10 T of cow. After using some of wheat and meat produced for wheat, meat production and feeding the labor, the farmer would like to save 10 T of wheat and 4 T of meat for his private consumption.
Model the production process of the farm using a system of linear equa- tions in the variables w (denoting the quantity of wheat), c (denoting the quantity of cow), and l (denoting the quantity of labor). Does the sys- tem of linear equations have a solution (where all the variables takes a non-negative value)?
As I remarked in the comments to your question, that last part isn't well defined, so I'll ignore it for now.
Your equations are \begin{align} -w=&\frac{1}{4}w+\frac{1}{10}c +\frac{1}{5}l\\ -c=&2w+\frac{1}{2}c+4l\\ -l=&\frac{1}{4}w+\frac{1}{10}c \end{align}
We rewrite these to give zero on the LHS and write them in the basis $\{w,c,l\}$: $$\mathbf{0}= \pmatrix{\frac{5}{4} & \frac{1}{10} & \frac{1}{5} \\ 2 & \frac{3}{2} & 4 \\ \frac{1}{4} & \frac{1}{10} & 1}$$
Can you solve it from here?
EDIT: I had forgotten the minus signs. To understand why, multiply by $-1$ on both sides of all equations and interpret it as: "To make one of this thing (plus) I need to use (minus) some of the others."