Hi, there--

THE PROBLEM:
Ira has 28¢ in his pocket. He has 20 coins that are either pennies or nickels. How many nickels does Ira have?

A SOLUTION:
Let N be the number of nickels in Ira's pocket.
There are 20 coins altogether, so the number of pennies is 20-N [because N + (20 - N) = 20].

The value of the nickels is 5¢ times the number of nickels, or 5N.
The value of the pennies is 1¢ times the number of pennies, or 1*(20-N)=20-N.

Ira has 28¢ in his pocket, so the value of the nickels and pennies is 28.
5N + 20 - N = 28.

Solve this equation for N. Combine like terms (5N-N=4N).
4N + 20 = 28

Subtract 20 from both sides of the equation.
4N + 20 - 20 = 28 - 20

4N = 8

Divide both sides of the equation by 4.
4N/4 = 8/4
N = 2

In this problem, N = 2 means that Ira has 2 nickels in his pocket.


Check your answer using the original words of the problem.
Since there are 20 coins altogether, there must be 20-2=18 pennies.
Two nickels are worth 10¢, and eighteen pennies are worth, 18¢. 
Ira has 10+18=28¢ in his pocket.


Hope this helps! Feel free to email if you have any questions about the solution.

Good luck with your math,

Mrs. F
math.in.the.vortex@gmail.com