Question 1065219

Franklin has $3.95 in nickels, dimes, and quarters. If he has three times as many dimes as quarters, and five fewer nickels than dimes, how many coins of each type does he have?

All I have is: 

3d=q
d=n-5
<pre>Your equations are incorrect!
Let the number of quarters be Q
Then number of dimes = 3Q
Number of nickels = 3Q - 5
We then get: .25Q + .1(3Q) + .05(3Q - 5) = 3.95
.25Q + .3Q + .15Q - .25 = 3.95
.7Q = 3.95 + .25
.7Q = 4.2
Q, or number of quarters = {{{highlight_green(matrix(1,3, 4.2/.7, or, 6))}}}
Number of dimes: {{{highlight_green(matrix(1,3, 3(6), or, 18))}}}
Number of nickels: {{{highlight_green(matrix(1,5, 3(6) - 5, or, 18 - 5, or, 13))}}}