Step-by-step explanation: I'll assume that the price of each has to be a whole cent value. Like, I can't have something worth 3 and 1/2 cents or something at the end.
If I assume the pencils are worth 1 cent then the three notebooks will have to be worth 8.46 or 2.82 apiece.
If I assume the pencils are worth 2 cents, then the three notebooks are worth 8.42 which isn't divisible by 3.
Pencils, 3 cents - notebooks, 8.38, not divisible by 3.
(you can tell very quickly whether a number is divisible by 3 by adding its digits and seeing if you get a multiple of 3)
Pencils, 4 cents - notebooks, 8.34, which is divisible by 3, so each notebook is worth 2.78.
It turns out that all I have to do is make a list where the 8.50 decreases by 4 each time. I'll find that every third answer I get, starting with the first, will be something I can divide by 3.
So I'll make a list of 8.46 - 0.12k and divide each of those by 3.
That's the same as 2.82 - 0.04k.
You get a list of the possible prices of the notebooks by letting k go between 0 and 70.
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If we had any more information, this could be narrowed down more (like if you had a second equation relating some number of pencils and notebooks...?).
Like, if we assumed that no one sells anything for less than 10 cents these days, and that the pencils are worth some multiple of 10 cents, we could start by doing the same thing for the pencils in increments of 10.
A pencil is worth 0.10, so that leaves 8.10, meaning each notebook is 2.70.
A pencil is worth 0.20, so that leaves 7.70, not divisible by 3.
A pencil is worth 0.30, so that leaves 7.30, not divisible by 3.
A pencil is worth 0.40, so that leaves 6.90, meaning each notebook is worth 2.30.
And we quickly see that the possible prices of the notebooks are 2.70 - 0.40k where k varies between 0 and 6. We only get 7 possibilities compared to the over 70 from before.
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Maybe we could assume that everything is some multiple of 25 cents: then we would get
A pencil is worth 0.25, so that leaves 7.50, which is divisible by 3 to get a multiple of quarters: each notebook being worth 2.50.
A pencil is worth 0.50, so that leaves 6.50, not divisible by 3.
A pencil is worth 0.75, so that leaves 5.50, also not divisible by 3.
A pencil is worth 1.00, so that leaves 4.50, which is divisible by 3 to get that each notebook is worth 1.50.
The next solution we would get would be a pencil being worth 1.75 and a notebook only being worth 0.50.
If we set the condition that a notebook had to be more than a pencil we'd get two solutions. If we set the conditions that a notebook had to be at least a dollar more than a pencil, we'd get one solution.
You can't narrow things down without more information.
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Does this make sense? Unless you have some way of fixing a single solution, you'll often get more than one possibility. All the possibilities from the second two scenarios should be contained within the possibilities of the first, but we had to pretend to have more information to single them out.
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Here's a pretend problem:
The cost of 3 notebooks and 4 pencils is $8.50 and the price of a notebook is ten times the price of a pencil.
You could set up two equations from this,
3N + 4P = 8.50
N = 10P
Where you're letting N and P represent the price of the notebooks and pencils.
Since you know that N is the same thing as 10P, you can rewrite the first equation as
3(10P) + 4P = 8.50
34P = 8.50
P = 0.25
Since the cost of a notebook is ten times this, N = 2.50.
Here, that second equation made it so we could find a single solution.
