There are three more nickels than quarters and six less dimes than quarters. The coins value $3.15. How many of each coin?

Question

how do i write an equation for this and solve it?

Answer 1

Here's what we know:

n = q + 3 (there are 3 more nickels than quarters)

q = d + 6 (6 less dimes than quarters also means there're 6 more quarters than dimes)

.05n + .1d + .25q = 3.15 (the total coinage value is $3.15)

Let's start plugging in values until we only have one variable left:

.05(q + 3) + .1d + .25(d + 6) = 3.15

.05((d + 6) + 3) + .1d + .25(d + 6) = 3.15

Simplify the parentheses:

.05(d + 9) + .1d + .25(d + 6) = 3.15

Distribute:

.05d + .45 + .1d + .25d + 1.5 = 3.15

Simplify by combining like terms:

.4d + 1.95 = 3.15

Subtract 1.95 from both sides

.4d = 1.2

d = 3

Now that we know the number of dimes, we can find the others:

q = 3 + 6; q = 9

n = 9 + 3; n = 12

check:

.05 * 12 + .1 * 3 + .25 * 9 = 3.15

.6 + .3 + 2.25 = 3.15

2.25 + .9 = 3.15

3.15 = 3.15