Question 1200476
<font color=black size=3>
Answer: <font color=red>21 nickels, 42 dimes, 13 quarters</font>



Work Shown:
n = number of nickels
d = number of dimes
q = number of quarters
n+d+q = 76 coins total
5n+10d+25q = 850 cents in total value
d = 2n since there are twice as many dimes as nickels


Applying substitution and solving for q.
n+d+q = 76
n+2n+q = 76 ... plug in d = 2n
3n+q = 76
q = 76-3n


Also,
5n+10d+25q = 850
5n+10(2n)+25q = 850 ... plug in d = 2n
5n+20n+25q = 850
25n+25q = 850
25(n+q) = 850
n+q = 850/25
n+q = 34
n+76-3n = 34 ... plug in q = 76-3n; solve for n
-2n+76 = 34
-2n = 34-76
-2n = -42
n = -42/(-2)
n = 21


Use that value to find the others.
d = 2n = 2*21 = 42
q = 76-3n = 76-3*21 = 76-63 = 13


In summary,
<font color=red>n = 21
d = 42
q = 13</font>
to represent the number of nickels, dimes, and quarters respectively.


Check:<ul><li>Coin count: 21 nickels + 42 dimes + 13 quarters = 21+42+13 = 76 coins total</li><li>Coin value: 21 nickels + 42 dimes + 13 quarters = 21*5 + 42*10 + 13*25 = 850 cents = $8.50 total value</li><li>The number of dimes (42) is double that of the number of nickels (21).</li></ul>The answers have been confirmed.


Confirmation using WolframAlpha
<a href = "https://www.wolframalpha.com/input?i=n%2Bd%2Bq+%3D+76%2C5n%2B10d%2B25q%3D850%2Cd%3D2n">https://www.wolframalpha.com/input?i=n%2Bd%2Bq+%3D+76%2C5n%2B10d%2B25q%3D850%2Cd%3D2n</a>
The CAS mode in GeoGebra has a similar solver feature.
</font>