SOLUTION: Suppose you have 12 coins that total 32 cents. Some coins are nickels, while the rest are pennies. How many of each coin do you have? Please show work and explain! Thank yo

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Suppose you have 12 coins that total 32 cents. Some coins are nickels, while the rest are pennies. How many of each coin do you have? Please show work and explain! Thank yo      Log On


   

Question 433133: Suppose you have 12 coins that total 32 cents. Some coins are nickels, while the rest are pennies. How many of each coin do you have?
Please show work and explain!
Thank you :)
Found 2 solutions by katealdridge, rwm:
Answer by katealdridge(100) About Me  (Show Source):
You can put this solution on YOUR website!
This is a system of equations
One equation involves the number of coins:
n + p = 12
The involves the total value:
.05n + .01p = .32
Solve the first equation for either variable, we'll choose n.
n%2Bp=12
n=12-p
Then take 12-p and substitute it in for n in the second equation.
.05%2812-p%29%2B.01p=.32
Then solve for p.
.60-.05p%2B.01p=.32
.60-.04p=.32
-.04p=-.28
p=7 So there are 7 pennies. This leaves 5 nickels.

Answer by rwm(914) About Me  (Show Source):
You can put this solution on YOUR website!
n+p=12
5n+p=32
n=5 p=7
5 nickels and 7 pennies
This problem was already solved
http://www.algebra.com/algebra/homework/Human-and-algebraic-language/Human-and-algebraic-language.faq.question.147633.html