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

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Question 147633: Suppose you have 12 coins that total 32 cents. Some of the coins are nickels and the rest are pennies. How many of each coin do you have?
Found 2 solutions by vleith, ptaylor:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Let p be the pennies and n be nickles
p+%2B+n+=+12
p+%2B+5n+=+32
Use elmination, subtract these two equations
-4n+=+-+20
n+=+5
So you have 5 nickles and 12-5=7 pennies

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
One approach:
Let x=number of nickels---------------5x=amount of nickels
Then 12-x=number of pennies---1(12-x)=amount of pennies
And we are told that the above amounts totals 32 cents so our equation to solve:
5x+(12-x)=32 or
5x+12-x=32 subtract 12 from each side
5x+12-12-x=32-12 collect like terms
4x=20 divide each side by 4
x=5------------------------------------number of nickels
12-x=12-5=7-------------------------------number of pennies
CK
5*5+7*1=32
25+7=32
32=32
Another approach:
Let x=number of nickels
And let y=number of pennies
Now we are told the following
x+y=12------------------------------------eq1
5x+y=32------------------------------------eq2
subtract eq1 from eq2 and we get:
4x=20
etc.

Hope this helps---ptaylor