- anonymous

Mixture Problem
We didn't go over them in class but they are on the online homework =(

- chestercat

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- anonymous

##### 1 Attachment

- zepdrix

Let P=Pounds of Peanuts, C=Pounds of Cashews, R=Pounds of Raisins.
Then we can establish an equation based on the first sentence they gave us,\[\large\rm P+C+R=9\]Some amount of Peanuts, Cashews and Raisins give us a total of 9 pounds, ya? :)

- anonymous

Okay, that makes sense so far =)

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## More answers

- zepdrix

We're going to set up another equation, this one is a little trickier.
This one is based on prices.

- zepdrix

The peanuts cost 1.50 per pound.
So the total price we pay for ALL of our peanuts is \(\large\rm 1.50P\).
Where P is the number of pounds of peanuts.

- zepdrix

Similarly, our cashews will have a total price of \(\large\rm 2.00C\)
While our raisins will total \(\large\rm 1.00r\)

- ♪Chibiterasu

WUMBO

- zepdrix

They tell us that the total costs, of all the nuts, is 13 dollars.
So when we total up all of these totals:\[\large\rm 1.5P+2C+1R=13\]I dropped any unnecessary 0's, hopefully that wasn't confusing.

- anonymous

All right, that makes sense =)

- zepdrix

We can actually set up one more equation! :O

- zepdrix

`twice as many peanuts` as `cashews`.
So the pounds of peanuts, P, should be twice as large as the pounds of cashews, C.

- zepdrix

We can write that relationship like this: \(\large\rm P=2C\)
Maybe read that as, "peanuts equals twice the cashews"

- zepdrix

We're going to go back and substitute this relationship into our other two equations.

- anonymous

Okay!

- zepdrix

Our weight equation:
\[\large\rm \color{orangered}{P}+C+R=9\]Will become\[\large\rm \color{orangered}{2C}+C+R=9\]
While our price equation:\[\large\rm 1.5\color{orangered}{P}+2C+1R=13\]will become\[\large\rm 1.5\color{orangered}{(2C)}+2C+1R=13\]

- anonymous

Makes sense
(Just as a side note, could this also be solved with a matrix?)

- zepdrix

Mmm definitely.
Lemme think..

- zepdrix

\[\large\rm \left(\begin{array}{ccc|c}1 & 1 & 1 & 9\\ 1.5 & 2 & 1 & 13\end{array}\right)\]Ah sorry for slow :) I'm not really familiar with augmented matrices in latex lol
Setting out our third column is a little tricky.
Since \(\large\rm P=2C\), subtracting 2C from each side gives us:
\(\large\rm 1P-2C+0R=0\)
So our matrix would be:\[\large\rm \left(\begin{array}{ccc|c}1 & 1 & 1 & 9\\ 1.5 & 2 & 1 & 13\\ 1 &-2 & 0 & 0\end{array}\right)\]

- zepdrix

Our third row* not third column, blah typo

- anonymous

Okay! So then the answer would be (4,2,3)?
And no worries! You saved my educational life! XP

- anonymous

Nah wait, that's not one of the answers lol.

- zepdrix

It's not?
Hmm, that sounds right :o

- anonymous

No wait, wrong problem! It is that. Lol sorry

- zepdrix

you silly billy -_-

- zepdrix

such a bum... just using a matrix calculator or something? lol

- anonymous

Yeah, a TI84.
Honestly the whole substitution thing loses me every time, not that your explanation was losing me! I was just curious about the matrix because I will probably use that on my test tomorrow.

- zepdrix

matrices are extremely useful in real world problems :)
So it's definitely not a waste of time to get comfortable with your calculator!

- anonymous

Cool beans! Thanks so much =)

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