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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? : 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?
Answer by vleith(1156) About Me  (Show Source):
You can put this solution on YOUR website!
Let p be the pennies and n be nickles
p + n = 12
p + 5n = 32
Use elmination, subtract these two equations
-4n = - 20
n = 5
So you have 5 nickles and 12-5=7 pennies
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? : 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?
Answer by ptaylor(1326) 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