Question 244834: A person has quarters, dimes, nickels,and pennies with a total value of $3.86. The number of nickels is twice the number of quarters. The number of quarters is two less than the number of dimes. There are 40 coins in all. Write and solve an equation to find the number of each coin. I have figured out that there are eight quarters, sixteen nickels, ten dimes, and six pennies. I can not figure out how to write an equation for this problem. Can you help me?
Found 2 solutions by jim_thompson5910, richwmiller:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let
p = # of pennies,
n = # of nickels
d = # of dimes
q = # of quarters
Since "There are 40 coins in all", this means that  (ie add up the individual coin counts to get the total). This is your first equation.
Because "The number of nickels is twice the number of quarters", we know that 
Since "The number of quarters is two less than the number of dimes", we also know that 
Finally, because "A person has quarters, dimes, nickels,and pennies with a total value of $3.86", we get the equation 
Remember that a penny is $0.01, a nickel is $0.05, a dime is $0.10, and a quarter is $0.25. If you multiply those individual values by their counts (the defined variables) and add them all up, you'll get the total coin value $3.86
So the fourth and final equation is
At the end of the translations, you get the four equations
From here, all you need to do is solve the system. There are plenty of options available, but I recommend using a calculator to set up a matrix to solve this problem.
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! we actually need for equations
n=2q
q=d-2
n=2(d-2)=2d-4
25q+10d+5n+1p=386
25(d-2)+10d+5(2d-4)+p=386
25d+10d+10d-50-20+p386
45d-70+p=386
p+q+d+n=40
p+(d-2)+d+2(d-2)=40
p+d-2+2d-4=40
p+4d-6=40
p+4d=46
p=46-4d
45d-70+p=386
45d+p=456
45d+46-4d=456
41d=410
d=10
q=d-2=8
n=16
p=40-10-16-8
p=40-34=6
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