SOLUTION: Suppose you have just enough money, in coins, to pay for a loaf of bread priced at $1.95. You have 12 coins, all quarters and dimes. Let q equal the number of quarters and d equal

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Question 1147868: Suppose you have just enough money, in coins, to pay for a loaf of bread priced at $1.95. You have 12 coins, all quarters and dimes. Let q equal the number of quarters and d equal the number of dimes. How many of each coin do you have?
I have the answer as 5 quarters and 7 dimes. I got it just by using common sense and did it in my head. Though, I would like to know how to actually solve it. All i’m really looking for is for the equation/s. From there on, I can figure out if it is substitution or elimination and solve it myself.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52798)   (Show Source):
You can put this solution on YOUR website!
.

The equation system you are asking for is d + q = 12 (1) (counting coins) 10d + 25q = 195 (2) (counting money, in cents). To solve the system, express d = 12-q from equation (1) and substitute it into equation (2). You will get 10*(12-q) + 25q = 195 120 - 10q + 25q = 195 15q = 195 - 120 15q = 75 q = 75/15 = 5. ANSWER. 5 quarters and 12-5 = 7 dimes.

Solved.

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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
    - Non-typical coin problems
    - 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.

A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.

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 greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

The important part of your question is about setting up the equations. You have two pieces of information:
(1) the total number of coins (quarters and dimes) is 12
(2) the value of the coins -- 25 cents for each quarter and 10 cents for each dime -- is $1.95, or 195 cents

If you (logically) use q for the number of quarters and d for the number of dimes, then

(1) the given number of coins is the sum of the number of quarters and the number of dimes:

(2) the value (in cents) of q quarters is 25*q or 25q; the value of d dimes is 10*d or 10d; the given total value of the coins is the sum of the values of the quarters and dimes:

Those are the equations shown by the other tutor, though not in exactly the same form.

You said you knew you could figure out, once you saw the equations, "...if it is substitution or elimination...". Note of course that you can always use either method.

The other tutor showed a solution using substitution, which would not be my choice. Whenever the two equations are in this form, I find elimination far easier; there is no need to solve for one variable and substitute.
25q + 10d = 195 q + d = 12 Multiply the the second equation by 10 and subtract from the first: 25q + 10d = 195 10q + 10d = 120 ----------------- 15q = 75 Then q = 75/15 = 5 d = 12-q = 7

But, again, the choice of method is yours....