SOLUTION: 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

(1)  nickels + dimes + quarters = 36      ( <<<---=== given )

     N       + D     + Q        = 36      ( the same )


(2)  N       + D   = Q - 2                ( <<<---=== given )


========>  Q + (Q-2) = 36

          2Q - 2 = 36  ====>  2Q = 36 + 2 = 38  ====>  Q = 19


So, we found number of quarters.  It is 19.

Next, you can reduce the problem from 3 unknowns to only two of them:


    we have 36-19 = 17 nickels and dimes, that are worth  6.15 - 19*0.25 = 1.40 dollars.


 N +   D =  17      (coins)    (1)
5N + 10D = 140      (cents)    (2)


Simplify

 N +   D =  17                 (1')
 N +  2D =  28                 (2')


Subtract eq(1') from eq(2').


      D = 28-17 = 11.


Answer.  11 dimes,  19 quarters  and  17-11 = 6 nickels.


Check.   5*6 + 11*10 + 19*25 = 615 cents.   ! Correct !

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)





.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)



6 + D = 17 ------ Substituting 6 for N in eq (v)