SOLUTION: I tried to find an equation with the the unknown amount of coins but could not find one. I believe I need 3 seperate equations but I am not sure. .05N+.10D=5.65 .05N+2 + .10D*2

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Question 149706: I tried to find an equation with the the unknown amount of coins but could not find one. I believe I need 3 seperate equations but I am not sure.
.05N+.10D=5.65
.05N+2 + .10D*2
(Not sure if this one is right. don't know the third one but I am pretty sure there is one Please help)

Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45. How many dimes does he have?
Found 3 solutions by stanbon, ikleyn, MathTherapy:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45. How many dimes does he have?
------------------------
Value EQUATION: 5N + 10D = 565
Value EQUAT: 5(N+8) + 10(2D) = 1045
----------------------------
Rearrange:
5N + 10D = 565
5N + 20D = 1045 - 80
----------------------------
Subtract 1st from 2nd to get:
10D = 400
D = 40 (# of dimes)
----------------
Substitute to solve for N:
5N + 400 = 565
5N = 165
N = 33 (# of nickels)
===========================
Cheers,
Stan H.
Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.
The system you are asking for is THIS :


5*N     + 10*D    =  565     cents     (1)
5*(N+8) + 10*(2D) = 1045     cents     (2)

One notice. 

     It is always more convenient to work with the integer number of cents than with decimal numbers of dollars.

-----------------------------

.
For coin problems and their detailed solutions see the lessons in this site:
    - Coin problems
    - More Coin problems
    - Solving coin problems without using equations
    - Kevin and Randy Muise have a jar containing coins
    - Typical coin problems from the archive
    - Three methods for solving standard (typical) coin word problems
    - More complicated coin problems
    - Solving coin problems mentally by grouping without using equations
    - Santa Claus helps solving coin problem

You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.

Read them attentively and become an expert in this field.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Coin problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.


Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
I tried to find an equation with the the unknown amount of coins but could not find one. I believe I need 3 seperate equations but I am not sure.
.05N+.10D=5.65
.05N+2 + .10D*2
(Not sure if this one is right. don't know the third one but I am pretty sure there is one Please help)

Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45. How many dimes does he have?
Your 1st equation: .05N + .1D = 5.65 is CORRECT, but you don't have a 2nd equation. Look below to see the 2nd equation you need to have.



.05N + .1D = 5.65 ------- eq (i)



.05(N + 8) + .1(2D) = 10.45

.05N + .4 + .2D = 10.45

.05N + .2D = 10.05 ------ eq (ii)



       .1D = 4.4 ------ Subtracting eq (i) from eq (ii)

         D, or number of dimes =  

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