Lesson Mixture: Two-Part, price or cost, both material amounts unknown

Source code of 'Mixture: Two-Part, price or cost, both material amounts unknown'

This Lesson (Mixture: Two-Part, price or cost, both material amounts unknown) was created by by josgarithmetic(39617)

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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.

Mixture:  Two-Part, price or cost, both material amounts unknown

This kind of mixture problem works just like a percent mixture two-part problem and the same variables can be assigned.  The only difference is that the concentration unit for price or cost will be MONEY per ITEM QUANTITY instead of PARTS per HUNDRED.

Examples of this kind of problem are like these:
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A coffee store merchant wants to make 70 pounds of a blend of coffee  having a price 8.05 dollars per pound from one at a 6.00 dollars per pound and another at a price of 10.25 dollars per pound.  How much of each coffee should be used?
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A candy company plans to make a mixture of candy soft, individually wrapped pieces.  Chocolate roll candies priced at 3.00 dollars per pound and nut-toffee pieces priced at 6.35 dollars per pound will be used.  How many pounds of each of these candies should be used for 190 pounds of mixture for a price of 4.70 dollars per pound?

Those two are really the same problem, but just different examples.    They can be analyzed and solved using variable assignments like what is done for percent-mixture problems.

ASSIGN VARIABLES
L = the low price for the less-expensive material
T = mixture price, or target price
H = the high price for the more expensive material
M = amount of the mixture, in whatever mass or count units
u = amount of the low priced, L material
v = amount of the high priced, H material

SYMBOLIC STATEMENTS
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{{{L<T<H}}}
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{{{(Lu+Hv)/M=T}}}
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{{{u+v=M}}}
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The T and the M equations are the system of equations to solve, for the unknown variables, u and v.

SOLVE SYSTEM FOR u AND v
'
{{{highlight_green((Lu+Hv)/M=T)}}} and {{{highlight_green(u+v=M)}}}

The "M" equation can be written as {{{v=M-u}}}.
{{{Lu+Hv=TM}}}
{{{Lu+H(M-u)=TM}}}
{{{Lu+HM-Hu=TM}}}
{{{Lu-Hu=TM-HM}}}
{{{u(L-H)=M(T-H)}}}
{{{u=M(T-H)/(L-H)}}}
{{{u=M((T-H)/(L-H))(-1/-1)}}}
{{{highlight(u=M(H-T)/(H-L))}}}

Return to the material sum, the M equation, to finish v.
{{{v=M-u}}}
Substituting the now found formula for u,
{{{v=M-M(H-T)/(H-L)}}}
{{{v=M(1-(H-T)/(H-L))}}}
{{{v=M(H-L-(H-T))/(H-L)}}}
{{{v=M(H-L-H+T)/(H-L)}}}
{{{highlight(v=M(T-L)/(H-L))}}}

Those are the solutions for the amount, u,  of the low price material, and the amount, v,  of the higher price material.   Other than examples for unknowns being u and v, any two variables in the system may be unknown and suitable algebraic steps taken to find symbolic solutions for them.