linear equation with 2 variables

[hortabise]

Jones spent 86 dollars on dictionaries and alamanacs. The cost of 1 dictionary and 1 almanac is 32 dollars. He needs to purchase 50% more almanacs than dictionaries. How many of each will he purchase and what is the cost of each book?

 

[JeffM]

Jones spent 86 dollars on dictionaries and alamanacs. The cost of 1 dictionary and 1 almanac is 32 dollars. He needs to purchase 50% more almanacs than dictionaries. How many of each will he purchase and what is the cost of each book?

This IS a hard problem. I am counting FOUR variables, not 2, and only 3 equations. But there is implied information that does not represent equations.

First NAME the relevant variables along with a concise definition. I'll start.

x = number of dictionaries

y = number of almanacs

a = ?

b = ?

What are the three equations?

 

[mmm4444bot]

I am counting FOUR variables

I'm not sure I agree. I see x and y as variables, but I see a and b as fixed parameters in this exercise (constants). Definitely four unknowns to determine, though.

 

[HallsofIvy]

I started to say that myself but there are four unknowns: x= the number of dictionaries, y= the number of almanacs, a= the price of a dictionary in dollars, and b= the price of an almanac in dollars.

Jones spent 86 dollars on dictionaries and alamanacs.

xa+ yb= 86.

The cost of 1 dictionary and 1 almanac is 32 dollars.

a+ b= 32.

He needs to purchase 50% more almanacs than dictionaries.

y= 1.5 x.

How many of each will he purchase

Find x and y.

and what is the cost of each book?

Find a and b.

 

[JeffM]

Once we agree that there are four unknowns to find, I am far from sure that it makes a big difference whether we say four variables or two variables and two parameters. Conceptually, I understand why it might be slightly better to go with two variables and two parameter: Jones is probably a price taker, not a price maker.

I'd really like to know what class the student is taking. If this is regular second year algebra, this problem is a lot harder than those I used to get in the 9th grade. Of course it is possible that the student forgot to give us the entire problem. Finding a solution to the problem as given is not difficult, but showing that it is unique (if it is) is a challenge that I shall forego.

 

[mmm4444bot]

I am far from sure that it makes a big difference whether we say four variables or two variables and two parameters.

I did not consider whether changing that terminology affects this particular exercise; I'm just speaking in general. (Does it have to make a big difference, or would making a little difference count just as much?) A quantity that does not vary within a given scenario should not be called variable. Some students have difficulty distinguishing the difference between these two concepts. I'm thinking that proper terminology would help in that regard. :cool:

This is an intriguing exercise, if it's complete.

 

[JeffM]

I did not consider whether changing that terminology affects this particular exercise; I'm just speaking in general. (Does it have to make a big difference, or would making a little difference count just as much?) A quantity that does not vary within a given scenario should not be called variable. Some students have difficulty distinguishing the difference between these two concepts. I'm thinking that proper terminology would help in that regard. :cool:

This is an intriguing exercise, if it's complete.

Upon reflection, this exercise seems to require the terminology of "unknown." Nothing is varying; the number of books purchased is just as constrained as the price of the books. I just picked up the term "variable" from the title of the thread and did not think about whether it was the most appropriate term. It is not; I erred.

This exercise is great if the students have been taught that: (1) the existence of a finite number of solutions to a system with n unknowns requires at least n independent and consistent constraints, but (2) the system has a unique solution if the n constraints are represented by n independent and consistent equations. This exercise is terrible if the students have been taught only that a system of n unknowns has a unique solution if it is described by n independent and consistent linear equations. The purpose of an exercise in high school math is to force students to master what they are being taught; it discourages them to be given problems that fit no pattern to which they have been exposed.

 

[mmm4444bot]

I just picked up the term "variable" from the title of the thread

And missed the 2

I'm not sure what you're thinking, when you say nothing is varying. Final answers are constants? Anyways, I'm done here, unless we get more info. Cheers

 

[JeffM]

And missed the 2

I'm not sure what you're thinking, when you say nothing is varying. Final answers are constants? Anyways, I'm done here, unless we get more info. Cheers

Mark

\(\displaystyle x + 2 = 5 \implies x = 3.\) In my opinion, x cannot vary; it is a specific number. The proper terms, I believe, are (1) "pronumeral" or (2) "unknown" rather than "variable" or "parameter."

y = 2x + 3. Here the proper terms are (1) "pronumerals" or (2) "variables" because we are not implying a specific number but a relationship that is true for various numbers.

y = m + bx. Here the proper terms are (1) "pronumerals" or (2) "variables and parameters" because we are implying a family of relationships and, within each member of the family, a specific relationship, defined by the parameters, applies to various numbers.

Just my opinion. And I too am done.

PS Glad to see that Denis's demonstration that the answer is unique.

 

[hortabise]

Equation with 2 variables

c = cost each dict; 32 - c = cost each almanac
n = number of dict; 1.5n = number of almanacs

nc + 1.5n(32 - c) = 86
Simplify:
c = 4(24n - 43) / n

Only n = 2 is possible; so do we call that unique?

Denis:
The correct answer is 2 dictionaries@$10 a piece and 3 almanacs@$22 a piece. You arrived at the correct answer, however how did you get it? I understand the first line which is nc + 1.5n(32-c) = 86, but I don't see how you arrive at
c = 4(24n-43)/n nor how you deduce from this point that n = 2. Could you please explain. Thank You