# Algebra

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Nadia has 32 coins made up of nickles, dimes, and quarters. The sum of the number of nickels and the number of quarters is three times the number of dimes. If the total value of the coins is \$4.60, how many of each kind does she have?
I can't figure out how to make this word problem into 3 equations.

• Algebra -

x nickels
y dimes and
z quarters:

equation base on number of coins
x + y + z = 32 ------ #1

equation based on value of coins

5x + 10y + 25z = 460, which reduces to
x + 2y + 5z = 92 ----- #2

"The sum of the number of nickels and the number of quarters is three times the number of dimes"
----> x + z = 3y

x -3y + z = 0 ----- #3

if you do #1 - #3 , you get
4y = 32
y = 8

so #1 becomes
x + 8 + z = 32
x + z = 24

#2 becomes
x + 16 + 5z = 92
x + 5z = 76

subtract those two ...
4z = 52
z = 13

back into the 1st
x + 8 + 13 = 32
x = 11

looks like 11 nickels, 8 dimes and 13 quarters

Check:
number of coins = 11+8+13 = 32
value of coins = 5(11) + 10(8) + 25(13) = 460
is 11+13 equal to 3 times 8 ??
yes!
All is good

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