problem with math question please help?

Question

You find 38 coins consisting only of nickels, dimes, and quarters, with a face value of $4.75. However, the coins all date from 1926, and are worth considerably more than their face value. A coin dealer offers you $8 for each nickel, $6 for each dime, and $22 for each quarter, for a total of $410. How many dimes did you find?

Answer 1

n + d + q = 38

5n + 10d + 25q = 475

8n + 6d + 22q = 410
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If we divide the 2nd eqn. by 5: n + 2d + 5q = 95

subtracting the first eqn: -n -d -q = -38
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We get d + 4q = 57 *******


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If we take the first eqn time -8 we get

-8n -8d -8q = -304 Adding the last eqn:
8n + 6d + 22q = 410

We get -2d + 14q = 106, so -d + 7q = 53

Add that to the starred eqn above: d + 4q = 57

And we get 11q = 110, so q = 10

Now, d + 4q = 57, so d + 4(10) = 57, so d=17.

17 dimes.

Answer 2

There are 38 coins,
The face value of the coins is $4.75
Let n, d, q stand for the no of the different coins.
So n +d + q = 38
and .05n + .1d + .25q = 4.75
You're getting $8 for each nickel, etc.
The final value of the coins is 410
So 8n + 6d +22q = 410
You have 3 eqs with 3 unknowns.
n + d + q = 38 *
5n +10d + 25q = 475 or n + 2d + 5q = 95 **
and 4n +3d +11q = 205 ***
I'll start you off on solving the system
Subt eq * from eq ** you get an eq in d and q
Subt 4 times * from *** and you get another eq in d and q.
Use the two eqs in d and q to find d and q. Go back to * to find n.
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