This is the last algebra problem, that I need help with.
Hank has a pocketful of nickels, dimes, and quarters. He has 18 coins in all worth $2.00. Find how many of each coin he has, if he has twice as many dimes as nickels.
Please, I have 2 more problems like this one, so I need you to be descriptive on the steps you took. So I don't have to ask anymore questions here,
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Verified answer
10 Dimes= 1.00
5 Nickels= .25
3 Quarters=.75
All together= 2.00
Let's say that N is the number of nickels,
D is the number of dimes and
Q is the number of quarters
Then,
N+D+Q=18 (18 coins altogether)
and each nickel is worth 5 cents, so the value of the nickels is 5 x N cents; each dime is worth 10 cents, so the value of the dimes is 10 X D cents and the value of the quarters is 25 X Q cents.
We then can make 3 different equations:
1. N+D+Q=18
2. 5N+10D+25Q=200(that's $2 in cents)
3. D=2N since there are twice as many dimes as nickels
You can solve them simultaneously by substituting 2N for D in equation #1 and making Q the subject (on the left hand side by itself) Q=18-3N
You then substitute 18-3N into equation #2 for the Q and also substitute 2N for D in equation #2:
5N + 20N + 25(18-3N) = 200
If you persist with solving this equation it will tell you how many nickels, then you can work out the rest. Good Luck.