SOLUTION: systems of equations: Larry has $4 in nickels and quarters. If there are 36 coins in all, how many nickels and how many quarters does Larry have.

Question 1082722: systems of equations: Larry has $4 in nickels and quarters. If there are 36 coins in all, how many nickels and how many quarters does Larry have.
Found 2 solutions by My Virtual Math Guide, ikleyn:
Answer by My Virtual Math Guide(2)   (Show Source): You can put this solution on YOUR website!
First, let's define variables. Let n = number of nickels, and q = number of quarters. If Larry has $4 in nickels and quarters, then our first equation is
0.05n + 0.25q = 4 (since a nickel is 5 cents, and a quarter is 25 cents)
Since there are 36 coins, then n + q = 36
Now, we have to solve the equation simultaneously. We can use the substitution method. We can rewrite the second equation as n = 36 - q, then substitute into the first equation to get:
0.05(36 - q) + 0.25q = 4
Simplify
1.8 - 0.05q + 0.25q = 4
Simplify further:
1.8 + 0.2q = 4
To find q, first subtract 1.8 from both sides of the equation to get:
0.2q = 4 - 1.8 = 2.2
Now, divide both sides by 0.2 to get:
q = 2.2/0.2 = 11
To solve for n, remember n + q = 36
So n + 11 = 36
subtract 11 from both sides of the equation:
n = 36 - 11 = 25
Now that we have the values of n and q, we can check our answers to make sure we are right.
Equation 1: 0.05n + 0.25q = 4
Let's substitute the values of n and q into the equation to see if it makes the equation true.
0.05(25) + 0.25(11) = 1.25 + 2.75 = 4
Our answer is correct.

Answer by ikleyn(53765)   (Show Source): You can put this solution on YOUR website!
.

N + Q = 36 (1) ("coins" equation) 5N + 25Q = 400 (2) (cents. "value" equation) To solve, express N = 36-Q from (1) and substitute into equation (2). You will get 5*(36-Q) + 25Q = 400, 180 - 5Q + 25Q = 400, 20Q = 400 - 180 ---> 20Q = 220 ---> Q = = 11. Answer. 11 quarters and 36-11 = 25 nickels. Check. 5*25 + 25*11 = 125 + 275 = 400 cents. Correct !

Solved.

The method I applied here, is called "the substitution method".

There is a bunch of lessons on coin problems
    - 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
    - More complicated coin problems
    - Solving coin problems mentally by grouping without using equations
    - Santa Claus helps solving coin problem
    - OVERVIEW of lessons on coin word problems
in this site.

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 and become an expert in solution of coin problems.

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".

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