Manipulating and converting mixture ratios I am unable to understand the following calculation for the total-fat to total-mixture ratio (adding the numerator and denominator values and taking the ratio of the two values). Also I am unable to understand what is going on in the last line, and how they have calculated the values as per the given requirements from the ratio. I am really unable to decipher this so I am requiring some help and guidance.

 A: Related questions:
if you mix in equal quantity drinks with alcohol content $40\%$ and $44\%$, what is the alcohol percentage of the mixture?
if you mix in equal quantity drinks with alcohol content $40\%$, $44\%$  and $51\%$, what is the alcohol content of the mixture?
in a drink with alcohol content $40\%$, what is the ratio of alcohol to water?
If you can reply to the above questions, you will understand your problem.
A: We are given that the three milks have fat:nonfat ratios $$4:5,\quad 5:6\;\;\text {and }\; 6:7,\tag{*}$$ and asked to find the fat:nonfat ratio of a mixture containing equal quantities of the three milks.

*

*Let's convert the milks' fat:nonfat ratios to a form in which
the milks' quantities (i.e., combined fat & nonfat contents) are
easier to compare and standardise: the respective
fat:TotalContents ratios are $$4:9,\quad 5:11\;\;\text {and }\; 6:13\tag1$$$$=572:1287,\quad 585:1287\;\;\text {and }\; 594:1287$$
$(1287$ is the lowest common multiple of $9, 11, 13).$
So, their fat:nonfat ratios are  $$572:715,\quad
585:702\;\;\text {and }\; 594:693.$$ So, a $1:1:1$ mixture of the
three milks has fat:nonfat ratio
$$(572+585+594):(715+702+693)$$$$=1751:2110.\tag2$$
Note that $(2)$ is just the weighted average
$$\frac{715}{2110}\left(\frac{572}{715}\right)+\frac{702}{2110}\left(\frac{585}{702}\right)+\frac{693}{2110}\left(\frac{594}{693}\right)$$
and alternatively derivable from the simple average of $(1):$
$$\frac13\left(\frac49+\frac5{11}+\frac6{13}\right)\;:\;1-\frac13\left(\frac49+\frac5{11}+\frac6{13}\right).$$
Note that taking neither the simple nor weighted average of $(*)$ results in the correct answer, because the three ratios' ratio units are not of the same size. To throw this into relief, here's an explicit formulation of the given problem:

*

*Given that $$(4+5)x=(5+6)y=(6+7)z,$$ evaluate
$$\frac{4x+5y+6z}{5x+6y+7z}.$$
To perform this evaluation, just express each of $y$ and $z$ in terms of $x,$ which will then get cancelled out to give the above numerical answer.
