Ratio and Proportion Concept clarity. 
The milk and water in two vessels $A$ and $B$ are in ratio of $4:3$ and $2:3 $ respectively. In what ratio should the liquids in both vessels be mixed to obtain a new mixture in vessel $C$ consisting of half milk and half water?

Solution:
Let $X$ be the amount of mixture taken from $A$.
Let $Y$ be the amount of mixture taken from $B$.
\begin{equation}
\frac{\frac{4x}{7}+\frac{2y}{5}}{\frac{3x}{7}+\frac{3y}{5}} = \frac{1}{1}       \tag{1}
\end{equation}
Now solving for $\dfrac{x}{y}$ we will get the solution.
But I got confused when I thought why not I should do:
Let $X$ be the total mixture from $A$.
Let $Y$ be the total mixture from $B$.
Then:
If total mixture of $A$ is $X$ then milk will be $4x$ and water will be $3x$.
If total mixture of $B$ is $Y$ then milk will be $2y$ and water will be $3y$. 
So if we add both the mixture, the resultant mixture should be $1/1$.
\begin{equation}
\frac{4x+2y}{3x+3y} =  \frac{1}{1}             \tag{2}
\end{equation}
So I unable to understand what is the difference between both the two equations ( (1) and (2) ) in terms of meaning ( results of $\dfrac{x}{y}$ from both the equations is obviously different ). 
As I am thinking that whether we take some part of both the mixture and mix it to get the $1/1$ resultant mixture or we take the both whole mixtures and mix it to get the $1/1$ resultant mixture, the resultant ratio (i.e $\dfrac{x}{y}$) of both the equations should be same. I know i am missing something but I am not getting it. I hope you understand my dilemma and help me through it.
 A: 
If total mixture (taken from) of A is x then milk will be 4x and water will be 3x.

No, that would claim $x=4x+3x$, since the amount of fluid equals the amount of milk plus the amount of water.
The fluid in A contains a $4:3$ milk to water ratio, so you are taking $\tfrac 47x$ of milk, and $\tfrac 37x$ of water, for a total of $x$.
So, you mixing $\tfrac 47x$ and $\tfrac 25y$ milk, and $\tfrac 37x$ and $\tfrac 35y$ water, each from A and B respectively.
The ratio of milk to water in the resulting mixture is: $(\tfrac 47 x+\tfrac 25y)\div(\tfrac 37x+\tfrac 35y)$, or $(20x+14y)\div(15x+21y)$.
We would like this ratio to equal $1/1$ (1:1 water to milk), so , we obtain $5x=7y$ or $x/y=7/5$ .. a 7:5 ratio of fluid from A to fluid from B.
If the containers are the same size, this means taking $\tfrac 7{12}$'s of A and $\tfrac 5{12}$'s of B.
A: A ratio how many parts of $A$ to how many parts of $B$.  Not how many parts of $A$ in the whole.
A mixture that is "half and half" means $1$ part (half of it) is milk and $1$ part (half of it) is water.  So the ratio is $1:1$.  Not $1:2$.  ($1:2$ would mean $1$ part mild to $2$ parts water.  So it would be $\frac 13$ mile and $\frac 23$ water.)
So if you have $x$ portions of $A$ and $y$ partions of $B$ then:
$A$ if $\frac 47$ milk so you have $\frac 47 y$ portions of milk from $A$.  And $B$ is $\frac 25$ milk so you have $\frac 25y$ portions of milk from $B$.  And in total you have $\frac 47x + \frac 25y$ portions of milk.
And the same for water: $A$ is $\frac 37$ water and $B$ is $\frac 35$ water so you have $\frac 37x + \frac 25y$ total portions of water.
And we want
$\frac {\frac 47x + \frac 25y}{\frac 37x + \frac 25y} = \frac 11$
Or $\frac 47x + \frac 25y = \frac 37x + \frac 25y$
.....

If total mixture of A is X then milk will be 4x and water will be 3x. If total mixture of B is Y then milk will be 2y and water will be 3y. 

Well, no,  $4x + 3x \ne X$.  If the total mixture from $A$ was $7X$ then milk will be $4x$ and water will be $3x$.  
Sin if the total mixture of $B$ was $5Y$ you'd get $2Y$ milk and $3Y$ water.
So you'd have $\frac {\frac 754x + 2y}{\frac 753x + 3y} = \frac 11$ or 
$\frac {4x + \frac 57 2y}{ 3x + \frac 57 3y} =\frac 11$.
Better yet if you took $35X$ from $A$ and $35Y$ from $B$ you would have $\frac {4*5x + 3*7Y}{3*5x + 2*7y} = \frac 11$.
Those will work.
A: Another way to look at it. 
$C$ will contain the total of $X+Y$ milk and water, a half of which must be milk. 
If $X$ amount is taken from $A$, then the amount of milk taken will be $\frac47 X$. 
If $Y$ amount is taken from $B$, then the amount of milk taken will be $\frac25 Y$.
Hence:
$$\begin{align}\frac47 X+\frac25Y&=\frac12(X+Y) \Rightarrow \\
\frac47X-\frac12X&=\frac12Y-\frac25 Y\Rightarrow \\
\frac1{14}X&=\frac1{10}Y \Rightarrow \\
\frac{X}{Y}&=\frac75.\end{align}$$ 
