SOLUTION: Jen Butler has been pricing​ Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $ 115.  Two adults and three children must pay $ 82

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Jen Butler has been pricing​ Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $ 115.  Two adults and three children must pay $ 82      Log On


   

Question 1136669: Jen Butler has been pricing​ Speed-Pass train fares for a group trip to New York. Three adults and four children must pay $ 115.  Two adults and three children must pay $ 82.  Find the price of the​ adult's ticket and the price of a​ child's ticket.
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
3a + 4c = 115    (1)

2a + 3c =  82    (2)


Apply the Elimination method. Multiply eq(1) by 2 (both sides); multiply eq(2) by 3. You will get


6a + 8c = 230    (3)

6a + 9c = 246    (4)


Subtract eq(3) from eq(4).  You will get


     9c - 8c = 246 - 230

      c      = 16.


Substitute c= 16 into eq(2)


     2a + 3*16 = 82

     2a = 82 - 48 = 34  =======>  a = 17.


Answer.  Adult ticket price is $17;  children ticket price is  $16.


CHECK.   3*17 + 4*16 = 115 dollars;  

         2*17 + 3*16 = 82  dollars.    ! Correct !

Solved.