Algebra 1 homework?

Question

Ok, I'm in 8th grade and in Algebra 1. I don't really think the should have put me in here without letting me go to Pre Alg. first because I'm not the worlds greatest mathmatician, but what ever heres the problem:
The local market is having a sale on whole fish. The prices are as follows:
You can buy an albacore and a barracuda for $21.
You can buy a barracuda and a carp for $24.
You can buy a carp and a dogfish for $32.
You can buy a dogfish and an eel for $37.
You can buy an eel and a flounder for $31.
You can buy a flounder and a gar for $25.
You can buy a gar and an albacore for $26.

Determine the cost of each individual fish.

Now I'm not asking you guys to solve it, just to tell me how to solve it, because to be honest I'm lost. lol First person who can help me solve it gets 10 points! Thanks in advance.

Answer 1

Step 1: add together all of those combinations. That gives you two of everything, so you can divide by 2 to work out the total cost of everything.

Step 2: We can choose 3 of the combinations so that we cover exactly 6 of the fish. We can work out the 7th by subtracting that from the total cost.
(Eg, (b and c) and (d and e) and (f and g) add up to 24 + 37 + 25, so a must be the difference between that and the total).

Once you've worked out one price, just step through each equation - if you know one, you can work out the price of the one it goes with. Or you could repeat step 2 for each fish.

Answer 2

This looks more like a logic puzzle than an algebra question.

Think about this: If you can buy an albacore and a barracuda for $21, or a barracuda and a carp for $24, then a carp must cost $3 more than an albacore.

Then if you can buy a carp and a dogfish for $32 or a dogfish and an eel for $37, the eel must cost $5 more than the carp, or $8 more than an albacore.

By the same reasoning, since you can buy an eel and a flounder for $31 or a flounder and a gar for $25, the gar must cost $6 less than the eel, or $2 more than an albacore.

Now we're getting somewhere! You can but a gar and an albacore for $26, and a gar costs $2 more than an albacore. So put that into algebraic terms, calling the price of an albacore A and the price of a gar G:

G = A + 2, G + A = 26. Substituting the first equation into the second, we get A + 2 + A = 26. Combining the A's and subtracting 2 from each side, we get 2A = 24. Dividing by 2 gives A = 12. An albacore costs $12.

Now that you know how much one fish costs, it should be pretty easy to figure out what the rest of them cost.

Answer 3

Another way to do this is to simply figure out the cost of every fish relative to another. For example, if an albacore and barracuda cost $21, and a barracuda and a carp cost $24, then the price of an albacore plus $3 is the price of a carp. If you want to simply guess a few times at the price of one fish after you have made a list of all of them relative to another, once you get one to work out you can use your list to find the rest.

Answer 4

Here's how you do it: make a table start like this:
Fishes | Prices
_____________________________________________
albacore AND barracuda
barracuda AND carp
carp AND dogfish
...
You see you align the matching fish with the line above;
and for each the price match; now you will see a pattern when you write down the last one. The rest is easy when you look at the table.

As for mathematician, I nearly failed Algebra I; but I studied until I was able to get a PhD in Physics by age 20. I am not smart, but I do know that you never quit and never put friends or a good time before studying. It is a hard world and the one thing that can save you and never be taken away is your education.