SOLUTION: I am trying to solve the following problem using a system of equations. 1. A collection of nickels and quarters is worth 4.65. There are 3 more nickels than quarters. Use a s

Algebra ->  Systems-of-equations -> SOLUTION: I am trying to solve the following problem using a system of equations. 1. A collection of nickels and quarters is worth 4.65. There are 3 more nickels than quarters. Use a s      Log On


   

Question 715170: I am trying to solve the following problem using a system of equations.
1. A collection of nickels and quarters is worth 4.65. There are 3 more nickels than quarters. Use a system of equations to determine how many coins are in the collection.
I have n+q=4.65, and (3+n(.05))+ q(.25)=total coins, but how do I solve this?
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Your error is in adding coins and getting dollars.
You can't do that.

A collection of nickels and quarters is worth 4.65. 
That translates as the MONEY equation

   ($.05)n + ($.25)q = $4.65

There are 3 more nickels than quarters. 
That translates as the COIN equation:

                  n = q + 3

So the system is

        .05n + .25q = 4.65
                  n = q + 3

You can simplify the first equation by clearing of decimals
by multiplying through by 100

           5n + 25q = 465
                  n = q + 3

Substitute q + 3 for n in the first equation:

           5n + 25q = 465
     5(q + 3) + 25q = 465
      5q + 15 + 25q = 465
           30q + 15 = 465
                30q = 450
                  q = 15

Substitute 15 for q in 
 
                  n = q + 3
                  n = 15 + 3
                  n = 18

So there were 18 nickels and 15 quarters.

Edwin