solving equations involving money

tvrn

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Ana has $.85 in nickles & dimes.She has 2 more nickles than dimes. How many nickles & dimes does she have?
 
Ana has $.85 in nickles & dimes.She has 2 more nickles than dimes. How many nickles & dimes does she have?

We aren't going to just "do" a problem for you. What have you tried? A good place to start is by defining variables to represent "unknown" quantities in the problem.

For example...you don't know the number of dimes, right?

let d = number of dimes she has

Then, the problem tells us that she has "2 more nickels than dimes," so we can conclude that
d + 2 = number of nickels

AND....would you agree that

money in dimes + money in nickels = total amount of money

Maybe it would be easier to work in cents here...she's got $0.85, which is 85 cents....

money in dimes + money in nickels = 85 cents

a dime is worth 10 cents. If she has "d" dimes, then the "money in dimes" would be 10 d......

10d + money in nickels = 85

Ok...you take it from here!
 
Ana has $.85 in nickles & dimes.She has 2 more nickles than dimes. How many nickles & dimes does she have?

Set up an equation with what you know.

The total = 0.85.

Nickle=0.05=N

Dime=0.1=D

Now because there are 2 more nickles than dimes, if 2 nickles are subtracted from the total, what's left will leave an amount which nickles and dimes will go into exactly the same number of times, in this case, our constant, lets call it A.

So,

0.85-2N = A(N+D)

Can you finish it from here?
 
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value of A that iam getting is 5. iam getting confused is it number of dimes or
nickles?
 
value of A that iam getting is 5. iam getting confused is it number of dimes or
nickles?

5 is correct.

Because we have taken into account the 2 extra nickels, the number of nickles and dimes left are equal, so there are 5 dimes and 5 nickels left.
 
Set up an equation with what you know.

The total = 0.85.

Nickle=0.05=N

Dime=0.01=D
This is very confusing. If "N" equals the number of nickels you have then it can't equal 0.05. You are confusing the number of nickles with the the value of a nickel.

Now because there are 2 more nickles than dimes, if 2 nickles are subtracted from the total, what's left will leave an amount which nickles and dimes will go into exactly the same number of times, in this case, our constant, lets call it A.
I have no clue why you would introduce a third unknown.

So,

0.85-2N = A(N+D)

Can you finish it from here?
 
This is very confusing. If "N" equals the number of nickels you have then it can't equal 0.05. You are confusing the number of nickles with the the value of a nickel.


I have no clue why you would introduce a third unknown.

Halls, maybe you haven't understood the way I explained it,

N=0.05, which is the value of the nickle, and is a known. I did not state that this is the number of nickles, or an unknown.

Same for dimes, when I say D, I mean 10 cents.

A is the number of dimes or nickels, since they are the same.

So, I created ONE equation with ONE unknown,A, because D and N are known.

0.85-2N = A(N+D)

0.85-2(0.05)=A(0.05+0.1)

0.75 = A(0.
15)

A = 5

So there are 5 nickles and 5 dimes.

This is what I think is algebraically the easiest way to do it, however, if you have an easier suggestion with N and D as the number I would like to see it as this is the way I would usually think. :)









 
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