Use an algebraic approach to solve the problem.

Question

Use an algebraic approach to solve the problem.

Suppose that Maria has 130 coins consisting of pennies, nickels, and dimes. The number of nickels she has is 30 less than twice the number of pennies; the number of dimes she has is 20 less than three times the number of pennies. How many coins of each kind does she have?

Answer 1

Let:

   P = number of pennies

   N = n umber of nickels

   D = number of dimes

 

Translate:

"130 coins consisting of pennies, nickels, and dimes"    means     P+N+D=130    [eq1]

 

"number of nickels she has is 30 less than twice the number of pennies"     means

                  N                   =   -30          +   2            *             P                       [eq2]

 

"number of dimes she has is 20 less than three times the number of pennies"     means

                   D                 =    -20         +   3     *                      P                        [eq3]

 

The math:

Notice that both the number of nickels and the number of times is stated in terms of the number of pennies.  So, we may substitute:

                P             +            N           +           D               =       130              [eq1 again]

                P             +     ( -30+2P)      +     ( -20+3P)         =       130

                     6P - 50 = 130

                      6P = 180

                        P = 30                [There are 30 pennies]

 

Now, use any of the equations to find the value of N and D:

     N = -30+2P                  [eq2 again]

     N = -30 + 2(30)

     N = 30                           [There are 30 nickels]

 

     D = -20+3P                    [eq3 again]

     D = -20 + 3(30)

     D = 70                         [There are 70 dimes]

 

Check:

   Is   30+30+70 = 130   ?

             130 = 130   ?yes

   Is    30 = -30 + 2(30)   ?

          30 = 30   ?yes

   Is    70 = -20 + 3(30)   ?

          70 = 70   ?  yes