Question

A pet store owner mixes two types of dog food costing $2.60 per pound and $3.80 per pound to make 40 pounds of a mixture costing $3.35 per pound. How many pounds of each kind of dog food are in the mixture?

Answer 1

lets introduce a variable, and call the number of pounds of $3.35 dog food \(x\) and so the number of pounds of $2.60 dog food is then \(40-x\) since the total is 40 pounds

Answer 2

i also made a mistake, since it is $3.80 and $2.60

Answer 3

so lets call the number of pounds of $3.80 dog food \(x\) and the number of pounds of $2.60 dog food \(40-x\)

Answer 4

the cost of \(x\) pounds of $3.80 dog food is \(3.80x\) and the cost of the rest is \(2.60(40-x)\)
for a total cost of
\[3.80x+2.60(40-x)\] you also know that the total cost is \(3.35\times 40=134\) so set them equal
\[3.80x+2.60(40-x)=134\] and solve for \(x\)

Answer 5

im still kind of confused, since I have to find out how many pounds of EACH kind of dog food are in the mixture. That just gives me one solution for x

Answer 6

I may just be reading this wrong but Im still confused sorry!

Answer 7

ok if you find x you certainly know the other one because they have to add up to 40 right?

Answer 8

first we find the number of pounds of the expensive food, then we find the number of pounds of cheaper food

Answer 9

we can solve it if you like
\[3.80x+2.60(40-x)=134\] maybe easier to start with
\[38x+26(40-x)=1340\] i.e. multiply by 10 to get rid of the annoying decimals,

Answer 10

duh. thanks I dont know why I asked lol

Answer 11

then multiply out to get
\[38x+1040-26x=1340\] combine like terms get
\[12x+1040=1340\] subtract 1040 to get
\[12x=300\] and finally divide by 12 to get
\[x=25\] now we recall that x was the expensive food, so the cheap one has to be 15 pounds

Answer 12

A much simpler setup would be:

1. Set up Variables:

x = 1st dog food type
y = 2nd dog food type

2. Set up Equations:

(Total Pound) x + y = 40
(Total Cost) 2.80x + 3.80y = 134

3. Solve Systems of Equations

After solving the system, you get:
x = 25
y = 15

Answer 13

@satellite73 continues to prefer the most confusing and least simplistic methods.

Answer 14

really? it sure looks like the above method is simpler than mine, but lets look closely
how does one solve
\[x+y=40\]
\[2.80x+3.80y=134\] after setting up the equation?

3. Solve Systems of Equations

but that is precisely where the work is, and precisely what i wrote above
what are the steps? well you can always use substitution
so you solve \(y=40-x\) which is what i wrote in line IN THE VERY FIRST LINE

so lets call the number of pounds of $3.80 dog food \(x\)and the number of pounds of $2.60 dog food \(40−x\)

then replace y in the second equation by \(40-x\) which is what i wrote in the third line \(3.80x+2.60(40−x)\) after explaining why, if the amount of expensive food is \(x\) then the amount of cheap food is \(40-x\)

so all my "confusing least simplistic" method is in fact explaining how and why to solve, which hero abbreviates into
3) solve the system

in other words, method is "confusing and least simplistic" because all you really have to say is "solve the problem"
sort of like saying, gee, his diagram for replacing the transmission in your car is so confusing. why doesn't he write

3) change the transmission

Answer 15

i would also like to add that
a) of course i have the utmost respect for hero and
b) the last time he commented on one of my posts it was taking me to task for not explainging how i got an equation

seems like one cannot win
i suggest that the only reason the "solve the system" method looks simpler is because it seems perhaps more familiar, but assuredly all the actual work is identical, and for identical reasons