It usually works well to follow directions. The model used in problem 1 can be used also for problem 2.
1a) c + a = 342 . . . . . . . the total number of tickets sold is 342
1b) 5c +12a = 2550 . . . total revenue was $2550
1c) (5c +12a) -5(c +a) = (2550) -5(342)
.. 7a = 840
.. a = 840/7 = 120
.. c = 342 -a = 222
2a) c +a = 764
2b) 2c +5a = 2992
2c) (2c +5a) -2(c +a) = (2992) -2(764)
.. 3a = 1464
.. a = 488
.. c = 764 -a = 276
276 children and 488 adults attended the event.
3) You can recognize this as a "sum and difference" problem. The difference between hikers (h) and bikers (b) is 178:
.. h -b = 178
Their total number is 676.
.. h +b = 676
This pair of equations (as with all sum and difference problems) is easily solved by elimination. Add the two equations.
.. 2h = 178 +676 = 854
.. h = 427
.. b = 427 -178 = 249
427 were hiking; 249 were riding their bikes.
