Question 716140
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Let *[tex \LARGE x] represent the number of pounds of less expensive coffee.  Let *[tex \LARGE y] represent the number of pounds of more expensive coffee.

The total number of pounds of the mixture is 30 lbs, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 30]


The cost of *[tex \LARGE x] pounds of $6.40 per pound coffee must be *[tex \LARGE 6.4x] and the cost of *[tex \LARGE y] pounds of $10.40 per pound coffee must be *[tex \LARGE 10.4y].  The cost of 30 pounds of $9.04 per pound coffee is *[tex \LARGE 9.04(30)\ =\ 271.20], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6.4x\ +\ 10.4y\ =\ 271.2]


Solve the 2X2 system for *[tex \LARGE x] and *[tex \LARGE y].  Personnally, I would use the substitution method, but then again all roads lead to Rome.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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