Lesson Mixing two concentrations for known amount of mixture

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About josgarithmetic: Academic and job experience with beginning & intermediate Algebra. Tutorial help mostly for Basic Math and up through intermediate algebra.

Very common as academic and as used in real situations, a certain quantity of mixture is wanted using some material of lower concentration to be mixed with some material of higher concentration to get a mixture of some desired target concentration. A farmer has two supplies of milk, each with a different percentage of milk fat and he wants some known amount of mixture of an intermediate percent milk fat. A lab technician has two different liquids with different concentrations of an acid and wants to mix the two to get some known amount of an intermediate concentration. Another person has two alloys of copper, each at different percents copper and wants a specific amount of an alloy with a certain intermediate percent of copper.

Those examples and so many like them are all the SAME problem, and instead of first using the actual values to find the amounts of materials to use, they can all be solved using only variables first. The given values can then be substituted at the end to find the resulting values. The question to be answered is how much of each material to mix to achieve the required amount of mixture at the required percentage or concentration.

The form is that given a high concentration material, H %, and a low concentration material, L %, how much of each must be mixed to get some M amount of a target concentration, T%. Variables may be chosen as unknowns, u amount of the L% concentration material and v amount of the H% concentration material.

A Summary of the Variables:
L = concentration of the low percent material
H = concentration of the high percent material
T = the target concentration wanted, percent
M = amount of the target mixture at desired T concentration
u = unknown amount of L material
v = unknown amount of the H material

EQUATIONS

Account for the concentration as Amount of pure material in Amount of Mixture.
The amount of pure material from u and v will be Lu%2BHv

;
The amount of mixture will be M.
The formula for the T percent result would be:
highlight%28%28Lu%2BHv%29%2FM=T%29

, and this is also linear.

Account for the materials used to get M amount of mixture.
highlight%28u%2Bv=M%29

We now, very generally, have a system of two equations in the two unknowns, u and v.
----------------------------------
highlight%28%28Lu%2BHv%29%2FM=T%29

----------------------------------
We need to solve this system for u and v. We can start by working with or simplifying the T equation, although exactly which steps to begin may be varied.

Lu%2BHv=MT

,
and taking the M equation, v=M-u

, so we can substitute,
Lu=MT-H%28M-u%29

.
We know that H%3ET

and that

, so instead of having the ratio of two negative numbers, we can simply multiply the right side by %28%28-1%29%2F%28-1%29%29

to obtain,
highlight%28u=M%28H-T%29%2F%28H-L%29%29

.

We also want a finished formula for v. This could, if we solved for u, simply use v=M-M%28H-T%29%2F%28H-L%29

; we can also adjust this formula for v:
v=M%281-%28H-T%29%2F%28H-L%29%29

v=M%28%28H-L%29%2F%28H-L%29-%28H-T%29%2F%28H-L%29%29

A SUMMARY OF THE UNKNOWN QUANTITIES OF MATERIALS FOR THE SYSTEM:
-------------------------------------
highlight%28u=M%28H-T%29%2F%28H-L%29%29

-------------------------------------

This formulation of solving works for any two-part mixture problem that fits this general form, although many exercise problems are designed so that much of the arithmetic steps are convenient to do without solving the system symbolically.

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