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

1. anonymous

2. 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? :)

3. anonymous

Okay, that makes sense so far =)

4. zepdrix

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

5. 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.

6. zepdrix

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

7. ♪Chibiterasu

WUMBO

8. 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.

9. anonymous

All right, that makes sense =)

10. zepdrix

We can actually set up one more equation! :O

11. zepdrix

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

12. zepdrix

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

13. zepdrix

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

14. anonymous

Okay!

15. 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$\large\rm 1.5\color{orangered}{(2C)}+2C+1R=13$

16. anonymous

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

17. zepdrix

Mmm definitely. Lemme think..

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

19. zepdrix

Our third row* not third column, blah typo

20. anonymous

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

21. anonymous

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

22. zepdrix

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

23. anonymous

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

24. zepdrix

you silly billy -_-

25. zepdrix

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

26. 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.

27. zepdrix

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

28. anonymous

Cool beans! Thanks so much =)