Answer:
20 nickels, 40 dimes, and 8 quarters
Step-by-step explanation:
This is a system of three equations with three unknowns. Let x = # of nickels, y = # of dimes, and z = # of quarters.
x + y + z = 68 (since there are 68 total coins)
.05x + .10y + .25z = 7
.10x + .05y + .25z = 6 (if the nickels were dimes and the dimes were nickels)
Multiply both sides of the last two equations by 100 to simplify, giving us:
5x + 10y + 25z = 700 or x + 2y + 5z = 140 if you divide both sides by 5.
10x + 5y + 25z = 600 OR 2x + y + 5z = 120 if you divide both sides by 5.
Let's solve the first equation for x. That gives us x = 68 - y - z. We can substitute this result into each of the final two equations to create a system of two equations with two unknowns, which is easier to solve. For the first equation:
(68 - y - z) + 2y + 5z = 140
68 + y + 4z = 140 or y + 4z = 72
For the second equation, we have:
2(68 - y - z) + y + 5z = 120
136 - 2y - 2z + y + 5z = 120
136 - y + 3z = 120
-y + 3z = -16
Now we have our system of two linear equations:
y + 4z = 72
-y + 3z = -16
If we add those two equations together, then we get 7z = 56 or z = 8. If z = 8, we can substitute this into either of the two previous equations. Let's choose the first. This gives us:
y + 4(8) = 72 or y + 32 = 72.
So y = 40. If y = 40 and z = 8, then we know:
x = 68 - y - z = 68 - 40 - 8 = 28 - 8 = 20.
Consequently, there were 20 nickels, 40 dimes, and 8 quarters in the piggy bank.
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