Jtdylktuy

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:

$$bgeq 0, ; sgeq 0,; fgeq 0$$

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?

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 $-8s +23cdot 900=18300$. Solving for $s$ yields $s=300$.

Plugging both $s$ and $f$ into the first equation then gives us $7b+6cdot 300+2cdot 900 =5700$ which results in $b=300$.

Hence, we have obtained the solutions $b=300, s=300, f=900$.