please help me
jason paid 16.88 for a 6-lb mixture of granola and dried apple chucks. If the granola cost 2.10 per pound and the dried apple chunks cost 4.24 per pound, how much pounds of each did he buy?

Question

please help me 
jason paid 16.88 for a 6-lb mixture of granola and dried apple chucks. If the granola cost 2.10 per pound and the dried apple chunks cost 4.24 per pound, how much pounds of each did he buy?

accessdenied · Answer

Write two equations that model the information given and then find the solution of them.
Consider these as variables:
$ g = \text{lbs of granola} \
d = \text{lbs of dried apple chunks}$

You need an equation for the total # of lbs and the total cost.

anonymous · Answer

so how would I write it out to solve it

accessdenied · Answer

Well, we have the unknowns specified.  We know that the sum of those should amount to the $6\ \text{lb} $ mixture we have.  We also know that we spent $$16.88$ total, where the granola costs $$2.10\ \text{per lb}$ and the apple chunks cost $$4.24\ \text{per lb}$.  The amount we spend for $g\ \text{lbs}$ of granola should then be $2.10g$ and the same done with $d\ \text{lbs}$ of dried apple chunks.

anonymous · Answer

8.44 pounds of each

accessdenied · Answer

8.44 would be much larger than the sum we were given, 6 lbs total.

accessdenied · Answer

Modelling the data given, I find the equations:
 $g + d = 6$, and
 $2.10g + 4.24d = 16.88$.
because "$g + d$" represents the total amount we bought ($6\ \text{lbs}$), and "$2.10g + 4.24d$" gives us the total / sum of what we spent, which should be $$16.88$.
Makes sense?

anonymous · Answer

iam trying to figure out the answer now cause iam still a little confuse

anonymous · Answer

i should let x be the part i dnt know right

accessdenied · Answer

Uhh, we can use any variable to represent what we don't know.  I used $g$ and $d$ because they just help remember what we're looking for while working.  You could also use $x$ and $y$ if you wanted to, though.

accessdenied · Answer

Like, if we used $x$ for $\text{lbs of granola}$ and $y$ for $\text{lbs of dried apple chunks}$, the equations would look like this:
  $x + y = 6$
  $2.10x + 4.24y = 16.88$

accessdenied · Answer

The variable is merely a representation of what we don't know and need to find.  The name we give it does not affect the problem, it's just there so we can actually play with the equation more easily than with stuff like boxes.  :)

anonymous · Answer

alright so the equation is 2.10+ 4.24 =16.88 but what about the 6lbs

anonymous · Answer

i understand what u r saying i jus dnt get how i suppose to set it up to solve

accessdenied · Answer

x + y = 6           < total lbs eq.
2.10x + 4.24y = 16.88     < total cost eq.

I think the easiest way to solve this system of equations is just substitution.
  x + y = 6
  x = 6 - y

2.10(6 - y) + 4.24y = 16.88

Like that (though it might help to multiply both sides of that by 100 to make the actual arithmetic easier since decimals are usually yucky)

anonymous · Answer

that seem so confuseing but i will try to solve it jus like that

accessdenied · Answer

Hmm.. could you explain whats confusing about it?  We're basically just using properties of equality (Adding stuff to both sides) so we can get a one-variable equation that allows us to solve for one variable, and then getting the other variable using the first variable we found.

anonymous · Answer

my answer dnt seem right I take 4.28 divide it into 6.34 which is .68

accessdenied · Answer

Could you try writing all the work?  I'll attempt to identify what went wrong.

anonymous · Answer

2.10(6 - y) + 4.24y = 16.88 I took 2.10 and times it by 6 which is 12.6 then 2.10 times y which is 2.10y so the new equation is 12.6-2.10y+4.24y=16.88 is that right so far

accessdenied · Answer

Yup, that's correct.

anonymous · Answer

then i put the variables togather 10y +4.24y=14.24 so now i got 12.6-14.24y=16.88 then i have to get the y by it self so i substracted 12.6 to both side which leaves 14.24y=4.28 then divid 14.24 to both side leaves y=3.33

anonymous · Answer

or is it jus y=3.

accessdenied · Answer

When you're putting the variables together, you should add together the terms with the "y".
  -2.10y + 4.24y = 2.14y

So: 12.6 + 2.14y = 16.88

anonymous · Answer

thank u

accessdenied · Answer

So, you should get y=2 with that
Then we have to go back to the equation with x and put in that y-value:
 x = 6-y   y=2
 x = 6-2
 x = 4

accessdenied · Answer

The moral of the story, I suppose, is that: Algebra is all about using relations of these symbols that represent values.  Once we have some relation (like, x+y=6), we can use those basic properties of Math like addition/multiplication/etc. to both sides
  For example, if we know that x+y=6, then we can subtract y from both sides because of addition property of equality.  x=6-y.
  When we know that second equation (2.10x + 4.24y = 16.88), we can also use that information about x=6-y since they're the $same$ variables, just in a different relation.

I hope that makes sense and I have helped you.  :)

accessdenied · Answer

Sorry, I should say, "subtraction property of equality", but it works the same.

anonymous · Answer

i had to read it a few times but i think i got it thanks

accessdenied · Answer

You're welcome.  It's a really basic idea that just goes so far in Mathematics in general (and usually stumps people easily, in my classes).  :P
Once you know the equation, its basically all just using properties to derive things.  Creating the equations just follows from stating the things you need to know, then figuring out how they are related in a problem.  :)

anonymous · Answer

lol

accessdenied · Answer

Not sure what the 'lol' is for.