The other tutor's answer is wrong.
d + n + p = 21
10d + 5n + p = 121
5n + p = 41
I tried to subtract the frist two equations first,
That was OK. You eliminated p by subtracting the 1st from the second:
10d + 5n + p = 121
-( d + n + p = 21)
---------------------
9d + 4n = 100
then work with the answer I got and tried to subtract the third one.
No that's no good because you eliminated p from the first two,
so you must also eliminate p from a different pair of equations
both of which have p in them. What you got does not contain p. So
you must first eliminate p from the 1st and 3rd, by subtracting the
3rd from the 1st:
d + n + p = 21
-( 5n + p = 41)
-------------------
d - 4n = -20
Now take what you got from subtracting the first two:
9d + 4n = 100
and put that with it and add the two equations:
9d + 4n = 100
d - 4n = -20
-----------------
10d = 80
d = 8
Now substitute 8 for d in
d - 4n = -20
8 - 4n = -20
-4n = -28
n = 7
Now substitute both d = 8 and n = 7 in one of the original equations,
say, the easiest one:
d + n + p = 21
8 + 7 + p = 21
15 + p = 21
p = 6
So the solution is (d,n,p) = (8,7,6)
Now we check:
8 dimes = 8 coins
7 nickels = 7 coins
6 pennies = 6 coins
-------------------
21 coins, that checks!
8 dimes = 80 cents
7 nickels = 35 cents
6 pennies = 6 cents
--------------------
121 cents or $1.21, that checks!
If you take the dimes away, you have only the 7 nickels
and the 6 pennies
7 nickels = 35 cents
6 pennies = 6 cents
--------------------
41 cents, that checks!
So the answer is correct.
Edwin
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