Keisha has 45 coins in her piggy bank, consisting of only nickels, dimes and quarters. She has one fewer nickel than she has dimes, and one fewer quarter than three times her number of nickels. How many nickels does she have?

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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