Question

Mixture Problem

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

Answer 1

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? :)

Answer 2

Okay, that makes sense so far =)

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

WUMBO

Answer 7

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.

Answer 8

All right, that makes sense =)

Answer 9

We can actually set up one more equation! :O

Answer 10

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

Answer 11

We can write that relationship like this: \(\large\rm P=2C\)

Maybe read that as, "peanuts equals twice the cashews"

Answer 12

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

Answer 13

Okay!

Answer 14

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\[\large\rm 1.5\color{orangered}{(2C)}+2C+1R=13\]

Answer 15

Makes sense

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

Answer 16

Mmm definitely.
Lemme think..

Answer 17

\[\large\rm \left(\begin{array}{ccc|c}1 & 1 & 1 & 9\\ 1.5 & 2 & 1 & 13\end{array}\right)\]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:\[\large\rm \left(\begin{array}{ccc|c}1 & 1 & 1 & 9\\ 1.5 & 2 & 1 & 13\\ 1 &-2 & 0 & 0\end{array}\right)\]

Answer 18

Our third row* not third column, blah typo

Answer 19

Okay! So then the answer would be (4,2,3)?

And no worries! You saved my educational life! XP

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

you silly billy -_-

Answer 24

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

Answer 25

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.

Answer 26

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

Answer 27

Cool beans! Thanks so much =)