SOLUTION: Pat has $3.80 in nickels and dimes. If there are 51 coins in all, how many of them are nickels

Question 518948: Pat has $3.80 in nickels and dimes. If there are 51 coins in all, how many of them are nickels

Found 2 solutions by Maths68, ankor@dixie-net.com:
Answer by Maths68(1474) (Show Source): You can put this solution on YOUR website!
Let
Nickels = x
Dimes = 51-x
Total Amount = $3.80 = 3.80*100=380 cents
5x+10(51-x)=380
5x+510-10x=380
5x-10x=380-510
-5x=-130
-5x/-5=-130/-5
x=26

Nickels = x = 26
Dimes = 51-x = 51-26 = 25

Check
========
5*26 + 10*25 = 380
130+250=380
380=380

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Pat has $3.80 in nickels and dimes.
Let's do this equation in cents
5n + 10d = 380
:
If there are 51 coins in all,
n + d = 51
Multiply by 10, subtract the 1st equation
10n + 10d = 510
5n + 10d = 380
----------------Subtraction eliminates d, find n;
5n = 130
n = 130/5
n = 26 nickels
then
51 - 26 = 25 dimes
;
:
Check solutions by finding the actual value:
.05(26) + .10(25) =
1.30 + 2.50 = 3.80
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