Let x = nickels , y = dimes, z = quarters.
x + y + z = 50
.05x + .10y + .25z = 5.60 or 5x + 10y + 25z = 560
.10y = 5 (.05x) = .25x or 10y = 25x
If 10y = 25x, then y = 5/2 x
Plugging the above into the first equation yields:
x + 5/2 x + z = 50
7/2 x + z = 50 or 7x + 2z = 100 if we multiply both sides by 2.
We can divide the second equation by 5 to get:
x + 2y + 5z = 112 or by substitution
x + 2 (5/2) x + 5z = 112
x + 5x + 5z = 112
6x + 5z = 112
We have a system of linear equations with two variables that we can solve now:
7x + 2z = 100
6x + 5z = 112
23x = 276 (To get this, multiply the top equation by 5, the bottom one by 2, and then subtract the two from each other)
x = 12
If x = 12, then y = 5/2 (12) = 30
Finally, x +y + z = 50
12 + 30 + z = 50
42 + z = 50
z = 8
So there are 12 nickels, 30 dimes, and 8 quarters.
