Question 713402
You have a general question, someone mixes two kinds of materials.  One has a lower given strength and the other has a higher given strength.  How much of each material must be mixed to get a certain amount of mixture of a particular intermediate strength.


This "strength" used in that general question may be cost, price, money, percent, some other kind of concentration unit.  


Here deals with this grocer's problem:


ASSIGN VARIABLES TO ALL QUANTITIES
L, the lower concentration of available material, $5.00/kg
H, the higher concentration of available material, $5.80/kg
T, the target concentration of resulting mixture, $5.50/kg
x, the amount of lower concentrated material to use, unknown
y, the amount of higher concentrated materaial to use, unknown
M, amount of the resulting mixture, 40 kg.


{{{highlight((Lx+yH)/M=T)}}}

That can be partly changed by multiplying both sides by M to get  Lx+yH=TM.

The mass or material sum equation will also be used for completing the system.
{{{highlight(x+y=M)}}}

This one easily allows us to substitute for either x or for y in the rational equation to solve for just one variable and get its value.  


{{{(Lx+yH)/M=T}}}
{{{Lx+yH=TM}}}
{{{Lx+(M-x)H=TM}}}
{{{Lx+HM-Hx=TM}}}
{{{(L-H)x+HM=TM}}}
{{{(L-H)x=TM-HM}}}
{{{(L-H)x=(T-H)M}}}
{{{x=M(T-H)/(L-H)}}}, and since the two differences here would both be negative, 
we can multiply by (-1)/(-1),
{{{x=M(H-T)/(H-L)}}}


In brief summary, this general two part mixture problem can be well solved all symbolically using the above process, leading to the use of these two formulas:


{{{highlight(x=M(H-T)/(H-L))}}} and {{{highlight(y=M-x)}}}.
Substitute the given values to find x, and then find the value for y, using these two final formulas.