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?

When you mix up a m/n ratio of A/B to get C, then $m+n$ units of C will contain m units of A and n units of B, or $(m+n)C = mA+nB$. I prefer to express this as $C = \frac{m}{m+n}A + \frac{n}{m+n} B$. (Note that the sum of the fractions is exactly 1, so I can work with just the fraction of A, knowing that the fraction of B is just one minus A's fraction.)
This gives the two equations $C = \frac{1}{5}A + \frac{4}{5} B$, $D = \frac{4}{7}A + \frac{3}{7} B$
Now I mix up paint E with a fraction $x$ of C and fraction $1-x$ of D (that would be a ratio $\frac{x}{1-x}$ of C/D). This gives me $E = x C + (1-x)D$, and we also want $E=\frac{5}{11}A + \frac{6}{11} B$. Equating the two formula and and comparing the quantities of A, B will give me a formula for $x$. In fact, we need only look at the fraction of A in E: This gives the formula
$$x \frac{1}{5} + (1-x)\frac{4}{7} = \frac{5}{11}$$
Working through the tedious details gives $x = \frac{45}{143}$.
To validate the answer, we take
$$\frac{45}{143} C + \frac{98}{143}D = \frac{9(A+4B)}{143} C + \frac{14(4A+3B)}{143}D = \frac{5}{11} A + \frac{6}{11}B$$
