Suppose you have 12 coins that total 32 cents. Some of the coins are nickels and the rest are pen How many of each coin do you have?

Let #n# be the number of nickels you have, and #p# the number of pennies. It holds that:

#n + p = 12#, since the total amount of coins is #12#, some being nickels, and some pennies.

#5n + p = 32#, since each nickel is worth #5# cents, and each penny #1#.

Subtract the top equation from the bottom, to get:

#4n = 20 => n = 5#

Since you have #5# nickels, the rest are pennies, or #7# pennies.

This problem can be set up algebraically by using the value of the nickels plus the value of the pennies equal to the total value of 32 cents.

The value of the nickels is #5n# where #n# is the number of nickels

The value of the pennies is #1p# where #p# is the number of pennies

Therefore

#5n + 1p = 32#

We can now determine the number of pennies by using the possible number of nickels

#p = 32 - 5n#

#p = 32 - 5(0)# 0 nickels means 32 pennies
#p = 32#

#p = 32 - 5(1)# 1 nickels means 27 pennies
#p = 32 - 5#
#p = 27#

#p = 32 - 5(2)# 2 nickels means 22 pennies
#p = 32 - 10#
#p = 22#

#p = 32 - 5(3)# 3 nickels means 17 pennies
#p = 32 - 15#
#p = 17#

#p = 32 - 5(4)# 4 nickels means 12 pennies
#p = 32 - 20#
#p = 12#

#p = 32 - 5(5)# 5 nickels means 7 pennies
#p = 32 - 25#
#p = 7#

#p = 32 - 5(6)# 6 nickels means 2 pennies
#p = 32 - 30#
#p = 2#