There are five nickels.
Let <em>w</em> = the number of pennies; <em>x</em> = the number of nickels; <em>y</em> = the number of dimes; <em>z </em>= the number of quarters.
Then, we have three equations with four unknowns:
(1) <em>w</em> + <em>x = y
</em>
(2) 3<em>w</em> = 2<em>z</em>
(3) 2<em>y</em> = 3<em>z</em>
However, there is a fourth <em>unstated </em>condition: <em>w</em>, <em>x</em>, y, and <em>z</em> must all be integers.
Assume that z = 6 (to avoid fractions).
From Equation (2), w = 4.
From Equation (3), y = 9.
Insert the values for <em>w</em> and <em>y</em> into equation (1).
4 + <em>x</em> = 9
x = 9 – 4 = 5
Thus, there are 4 pennies, 5 nickels, 9 dimes, and 6 quarters.
<em>Check</em>: 4 pennies + 5 nickels = 9 dimes
6 quarters + 4 pennies = 3 quarters for every 2 pennies
9 dimes + 6 quarters = 3 dimes for every 2 quarters
