Mixing things- ratios Suppose I take two things, A and B.  C is made from a $1/4$ ratio of $A$ to $B$, while D is made from a $4/3$ ratio. If I want to know what ratio of $C$ to $D$ will give a $5/6$ ratio of $A$ to $B$, do I just solve the system 
$A+4B=C, 4A+3B=D, 5A+6B=z$
to get
$z= \frac{9C+14D}{13} \rightarrow \frac{5A}{6B} = \frac{9C}{14D}?$ If not what am I doing wrong? What is the right method?

The mixing ratios give
$$
\begin{bmatrix}
\frac15&\frac45\\
\frac47&\frac37
\end{bmatrix}
\begin{bmatrix}A\\B\end{bmatrix}=\begin{bmatrix}C\\D\end{bmatrix}
$$
That is, $1$ can of $C$ is $\frac15$ can of $A$ and $\frac45$ can of $B$, $1$ can of $D$ is $\frac47$ can of $A$ and $\frac37$ can of $B$.
Because
$$
\begin{align}
\begin{bmatrix}\frac5{11}&\frac6{11}\end{bmatrix}
\begin{bmatrix}
\frac15&\frac45\\
\frac47&\frac37
\end{bmatrix}^{-1}
&=\begin{bmatrix}\frac5{11}&\frac6{11}\end{bmatrix}
\begin{bmatrix}
-\frac{15}{13}&\frac{28}{13}\\
\frac{20}{13}&-\frac7{13}
\end{bmatrix}\\[6pt]
&=\begin{bmatrix}\frac{45}{143}&\frac{98}{143}\end{bmatrix}
\end{align}
$$
we get
$$
\begin{bmatrix}\frac5{11}&\frac6{11}\end{bmatrix}
\begin{bmatrix}A\\B\end{bmatrix}
=\begin{bmatrix}\frac{45}{143}&\frac{98}{143}\end{bmatrix}
\begin{bmatrix}C\\D\end{bmatrix}
$$
Therefore, a $45:98$ mixture of $C$ and $D$ requires a $5:6$ ratio of $A$ and $B$.
