Question 493919: I have 26 coins, worth $3.45, comprised of nickels, dimes, and quarters. The total number of dimes is four less than the combined total of nickels and quarters. How many of each do I have?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! I have 26 coins, worth $3.45, comprised of nickels, dimes, and quarters. The total number of dimes is four less than the combined total of nickels and quarters. How many of each do I have?
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let x= number of dimes
let y= number of nickels
let z= number of quarter
..
x+y+z=26
.1x+.05y+.25z=3.45
26-x= combined total of nickels and quarters
..
x=26-x-4
2x=22
x=11
..
x+y+z=26
11+y+z=26
y+z=15
..
.1x+.05y+.25z=3.45
sub 11 for x
1.1+.05y+.25z=3.45
.05y+.25z=2.35
multply by 100
5y+25z=235
y+z=15
multiply by 5
5y+5z=75
subtract from preceding equation,5y+25z=235
20z=160
z=8
y=15-8=7
x=11
..
Ans:
11 dimes
7 nickels
8 quarters
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