Can I buy 7 pencils and 7 notebooks for $15?

[rebo1984]

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

 

[MarkFL]

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. :)

 

[rebo1984]

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.

 

[MarkFL]

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?

 

[rebo1984]

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