Trail Mix Packaging problem part 1.1 I apologize in a advanced, I have tried looking at the other steps to what could be similar problems on stackexchange and could not understand the concepts.  I am trying to teach myself math from this book A first course in linear algebra and already stuck on the first question:
$$b\geq 0, \; s\geq 0,\; f\geq 0$$
Second, if we want to consume all of our ingredients each day, the storage capacities lead to three (linear) equations, one for each ingredient.
$$\frac{7}{15}b + \frac{6}{15}s + \frac{2}{15}f = 380 (raisins)$$
$$\frac{6}{15}b +\frac{4}{15}s + \frac{5}{15}f = 500 (peanuts)$$
$$\frac{2}{15}b + \frac{5}{15}s + \frac{8}{15}f = 620(chocolate)$$
Now the solution which is required, is: 
$$b=300 kg, s=300 kg, f=900 kg.$$
I have tried a few different concepts that I am too embarrassed to share as it's been 20 years since I studied math and don't remember ever doing this.  Would someone be able to give me a clue as to how to proceed?  As well, are there terms I can use to search on line for equations such as these to assist in my learning in the future?
 A: You can solve the problem with any algorithm for a system of linear equations. If you want to calculate the solution by hand then things will get somewhat easier if you multiply each equation with $15$ to remove the denominators:
$$7b + 6s + 2f = 5700 (raisins)$$
$$6b +4s + 5f = 7500 (peanuts)$$
$$2b + 5s + 8f = 9300(chocolate)$$
By adding the $-\frac{6}{7}$-multiple of the first equation to the second and the $-\frac{2}{7}$-multiple of the first equation to the third equation you obtain the following system:
$$7b + 6s + 2f = 5700 (raisins)$$
$$\frac{-8}{7}s + \frac{23}{7}f = \frac{18300}{7} (peanuts)$$
$$\frac{23}{7}s + \frac{52}{7}f = \frac{53700}{7}(chocolate)$$
Observe that now you have removed the $b$'s from the second and third equation. Multiplying the second and third equation by 7 will once more remove the denominators and we obtain
$$7b + 6s + 2f = 5700 (raisins)$$
$$-8s + 23f = 18300 (peanuts)$$
$$23s + 52f =53700(chocolate).$$
Now, we want to remove the $23s$ in the third equation. To do so, we add the $\frac{23}{8}$-th multiple of the second equation to the third. We obtain the system:
$$7b + 6s + 2f = 5700 (raisins)$$
$$-8s + 23f = 18300 (peanuts)$$
$$118.125f =106312.5(chocolate).$$
So, we know that $118.125f =106312.5$ which implies (divide both sides by 118.125) that $f=900$. By plugging this solution of $f$ into the second equation we get $-8s +23\cdot 900=18300$. Solving for $s$ yields $s=300$. 
Plugging both $s$ and $f$ into the first equation then gives us $7b+6\cdot 300+2\cdot 900 =5700$ which results in $b=300$.
Hence, we have obtained the solutions $b=300, s=300, f=900$.
