MHB Buy 7 Pencils & Notebooks with \$15 - Solve Inequalities

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The discussion revolves around how to allocate $15 to buy pencils and notebooks, with pencils priced at $0.50 each and notebooks at $1.50. An initial calculation suggested buying 7 of each, but it was pointed out that this would leave $1 unspent. The correct approach involves a Diophantine equation, leading to multiple combinations of pencils and notebooks that can be purchased while using the entire budget. Specifically, for each notebook removed from the cart, three pencils can be added, resulting in 11 valid combinations. Understanding the equations and their implications is crucial for solving the problem effectively.
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I have \$15 to buy pencils and notebooks from a bookstore. If pencils are
\$0.50 each and a notebook costs \$1.50, then how many pencils and
notebooks can I buy if I spent all of the money?

I reasoned: .5x+1.5x=15
x= 7.5
So 7 notebooks and 7 pencils.

Is this correct?Thanks
 
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But, you don't spend all of your money...you will have a dollar left over. So, you could by 9 pencils and 7 notebooks.

If you let $P$ be the number of pencils and $N$ be the number of notebooks, you obtain he Diophantine equation:

$$P+3N=30$$

This has the integer solutions:

$$P=3n$$

$$N=10-n$$

where $$0\le n\le10$$ and $n\in\mathbb{Z}$

This gives you 11 combinations that will work by letting $n$ rage over the integers from 0 to 10.

Another way to look at it is to observe that you could buy 10 notebooks for \$15. Then for every notebook you "put back" you can add 3 pencils to your cart. :)
 
MarkFL said:
But, you don't spend all of your money...you will have a dollar left over. So, you could by 9 pencils and 7 notebooks.

If you let $P$ be the number of pencils and $N$ be the number of notebooks, you obtain he Diophantine equation:

$$P+3N=30$$

This has the integer solutions:

$$P=3n$$

$$N=10-n$$

where $$0\le n\le10$$ and $n\in\mathbb{Z}$

This gives you 11 combinations that will work by letting $n$ rage over the integers from 0 to 10.

Another way to look at it is to observe that you could buy 10 notebooks for \$15. Then for every notebook you "put back" you can add 3 pencils to your cart. :)

Im not sure i understand how you derived those equations and what they mean.
 
rebo1984 said:
Im not sure i understand how you derived those equations and what they mean.

Perhaps you are not familiar with Diophantine Equations... (I didn't learn about them until I got into number theory), but focus instead on my analogy of putting 10 notebooks into your cart. That's one solution, and it will cost all of the \$15 you have. Now, for each notebook you put back, you can replace with 3 pencils, since 3 pencils cost the same as a notebook. So, the 11 possible solutions are:

$$(N,P)=(10,0),\,(9,3),\,(8,6),\,(7,9),\,(6,12),\,(5,15),\,(4,18),\,(3,21),\,(2,24),\,(1,27),\,(0,30)$$

Does that make sense?
 
Yes, that makes sense. And you're right: I'm not familiar with those equations.
 
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