Question 148577
hello, you know 1 nickel = 5 cent, 1 dime = 10 cent, 1 quarter = 25 cent.

   given there are 19 coins. so let us take the number of nickels  are N, dimes are D, and  quarters  are Q .

  i.e    N + D + Q = 19 .........(1)

 also  it is given there are two more dimes than nickels,
    i.e.         D = N + 2 ........(2)


    It is given the total value of all the coins is $2.35 = 235 cents

       i.e  N nickels + D dimes + Q quarters = 235 cents

             N * 5 + D * 10 + Q * 25 = 235 .........(3)   ( Substituting the
                                     values of nickel, dime  & quarter in cents)
substituting (2) in (1)we get,
 
            N + N+ 2 + Q = 19 

               2N + 2 + Q = 19 
                     2N + Q = 19 - 2 = 17 
                        Q = 17 - 2N   ............(4)
SUBSTITUTING (2)  & (4) in (3)

       N * 5 +  (N+2) * 10 + (17 - 2N) * 25 = 235 

         5N + 10N + 2*10 + 17*25 - 2N *25 = 235
             15N  + 20 + 425 - 50N = 235 

                  445 - 35N = 235 
                    445 - 235 = 35 N 
                       210    = 35 N 

                       N  = 210/ 35 = 6 
put N= 6  in (2)
               D = N + 2 = 6 + 2 = 8 .

put N=6 in (4) 

              Q = 17 - 2N = 17 - 2* 6 = 17- 12 = 5 .
i.e. mike  have 6 coins of nickel, 8 coins of dimes and 5 quarter coins .