Question 974752

a box contains $7.15 in nickels, dimes, and quaters,. There are 42 coins in all and the combined total of nickles and dimes is 2 less than the total number of quaters. Find the number of each type of coin, using systems of equations, and showing work.
<pre>Let number of nickels, dimes, and quarters, be N, D, and Q, respectively
Then we get: N + D + Q = 42 ------- eq (i)
Also, N + D = Q - 2_____N + D - Q = - 2 ------- eq (ii)
2Q = 44 ------ Subtracting eq (ii) from eq (i) 
Q, or number of quarters = {{{44/2}}}, or {{{highlight_green(22)}}}

N + D + 22 = 42 ------- Substituting 22 for Q in eq (i)
N + D = 20
N = 20 - D

Since all coins total $7.15, we then get:
.05N + .1D + .25(Q) = 7.15
.05(20 - D) + .1D + .25(22) = 7.15	
1 - .05D + .1D + 5.5 = 7.15	
- .05D + .1D + 1 + 5.5 = 7.15	
.05D + 6.5 = 7.15
.05D = .65
D, or number of dimes = {{{.65/.05}}}, or {{{highlight_green(13)}}}
N = 20 - D 
N = 20 - 13 ------ Substituting 13 for D
N, or number of nickels = {{{highlight_green(7)}}}