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If you are doing the same exercise, I would recommended viewing the Trail Mixed Packages problem part 1 and 2 before reading here, otherwise you are only cheating yourself.

Again, reflecting on the original question below which, can be found here, under the subtitle Example TMP: Trail Mixed Packaging combined with the help from these two sources: Damien & YukiJ

Excerpt from Example TMP: Trail Mixed Packaging.

Let us denote the amount of each mix to produce each day, measured in kilograms, by the variable quantities b, s and f. Your production schedule can be described as values of b, s and f that do several things. First, we cannot make negative quantities of each mix, so

$$b≥0s≥0f≥0$$.

Second, if we want to consume all of our ingredients each day, the storage capacities lead to three (linear) equations, one for each ingredient.

$$frac{7}{15}b+frac{6}{15}s+frac{2}{15}f=380(raisins)$$
$$frac{6}{15}b+frac{4}{15}s+frac{5}{15}f=500(peanuts)$$
$$frac{2}{15}b+frac{5}{15}s+frac{8}{15}f=620 (chocolate)$$

It happens that this system of three equations has just one solution. In other words, as production manager, your job is easy, since there is but one way to use up all of your raw ingredients making trail mix. This single solution is

$$b=300 kgs=300 kgf=900 kg.$$

Show of workings for TMP caculation:

The first part is straight forward, to solve everything just needs to be multiplied by 15.

$$7b+6s+2f=5700 (raisins)$$
$$ 6b+4s+5f=7500 (peanuts)$$
$$ 2b+5s+8f=9200 (chocolate)$$

To remove $b$ we add and multiple $-frac{6}{7}$ for the first and second equation. Then we add and mulitple the first and third equation by $-frac{2}{7}$ as follows:

$$Eq2. frac{6-6}{7}b+frac{4*7-6*6}{7}s+frac{5*7-6*2}{7}f=frac{7500*7-5700*6}{7}$$

$$Eq3. frac{2-2}{7}+frac{5*7-2*6}{7}+frac{8*7-2*2}{7}=frac{9200*7-5700*2}{7}$$

In the end we are left with the values for $s$ and $f$.

$$6s+2f=380 (raisins)$$
$$frac{-8}{7}s+frac{23}{7}f=18300 (peanuts)$$
$$frac{23}{7}s+frac{52}{7}f=53000 (chocolate)$$

Now, we remove the 7 by multiplying the third equation by 7 which, cancels out the the 7 in the second equation and vice versa.

$$6s+2f=380 (raisins)$$
$$-8s+23f=18300$$
$$23s+52f=53000$$

From here the 23s can be extracted by adding the multiple of $frac{23}{8}$.

$$frac{-23s+23}{8}s+frac{52*-8+(-23*23)}{8}f =frac{53000*8+23*18300}{8}$$

After multiplying $frac{-1}{1}$ by the result $-118.125f = -105612.5$ gives me:

$$118.125f = 105612.5$$

At this point in time, my question is this, why is it that I have different answers on the right side of the equation 2 and later on equation 3 compared to part 1.1 post? Secondly, I will try for the answer later on, just will have a break.

You made a really early unfortunate mistake. For the equation for chocolate, you originally had $$frac2{15}b+frac5{15}s+frac8{15}f=620$$
which you then turned into $$2b+5s+8f=9200$$by multiplying both sides by $15$. Actually, we have $620times15=9300$. If you correct this, you'll find that the final equation will be $$118.125f=106312.5$$ which does indeed give $f=900$.