Question 1114184
A box contains $6.15 in nickels , dimes, and quarters. There are 36 coins in all, and the sum of the numbers of nickels and dimes 
is 2 less than the number of quarters. How many coins of each kind are there? 
<pre>Let the number of nickels, dimes, and quarters be N, D, and Q, respectively
Then we get: .05N + .1D + .25Q = 6.15 ------ eq (i)
Also, N + D + Q = 36 ------- eq (ii)
In addition, N + D = Q - 2______N + D - Q = - 2 ------ eq (iii)
2Q = 38 ------ Subtracting eq (iii) from eq (ii)
{{{highlight_green(matrix(1,9, "Q,", or, number, of, quarters, "=", 38/2, "=", 19))}}}

.05N + .1D + .25(19) = 6.15 ------ Substituting 19 for Q in eq (i)
.05N + .1D + 4.75 = 6.15
.05N + .1D = 1.4 ------- eq (iv)

N + D = 19 - 2 ------- Substituting 19 for Q in eq (iii)
N + D = 17 ------ eq (v)
- .1N - .1D = - 1.7 ------- Multiplying eq (v) by - .1 ------ eq (vi)
- .05N = - .3 ------ Adding eqs (vi) & (iv)
{{{highlight_green(matrix(1,9, "N,", or, number, of, nickels, "=", (- .3)/(- .05), "=", 6))}}}
6 + D = 17 ------ Substituting 6 for N in eq (v)
{{{highlight_green(matrix(1,9, "D,", or, number, of, dimes, "=", 17 - 6, "=", 11))}}}