system of three linear equations

Question

a box contains $2 in nickels, dimes, and quarters. there are 19 coins in all, and there are twice as many nickels as dimes. how many coins of each kind are there?

Answer 1

This seems like a pretty simple algebra problem. 

 

Let's call the number of dimes "x." 

1. There are x dimes.
2. There are twice as many nickels as dimes, so that's 2x.
3. There are 19 coins in all, so the quarters are 19 - x (dimes) - 2x (nickels).
4. All the coins add up to $2.00, so if we multiply the various items by the values of the coins, we get $2.00.

 

200 (pennies) = 10x (dimes) +5 * 2x (nickels) + 25(19-x-2x) (quarters)

200 = 10x + 10x +25(19-3x)

200 = 20x +475 -75x

200 = 475 -55x

55x = 275

x = 5

 

Plugging it back in: 5 dimes ($.50) plus 10 nickels ($.50) plus 4 quarters ($1.00) = $2.00.