Question 915932
To solve, we need to set this word problem up as a system of linear equations.  First, let's use the letter A to represent the Adult tickets and C to represent the child tickets.  The problem says that 3 Adults and 4 children must pay 129 dollars.  So,


3A + 4C = 129


Next, we are told that 2 Adults and 3 children must pay 92 dollars.  So,


2A + 3C = 92


Our system of equations are


3A + 4C = 129
2A + 3C = 92


We need to now solve for either A or C.  Let's choose C.  To solve for C, we need to multiply the first equation by a certain number, so that when we add that first equation to the second equation, our letter A will disappear.  If we multiply equation 1 by -2 and the second equation by 3 and then add them together, our A will go away:


-2(3A + 4C = 129) -----> -6A - 8C = -258


3(2A + 3C = 92) -----> 6A + 9C = 276


Now, let's add these two results together:


-6A - 8C + 6A + 9C = 276 - 258 -----> C = 18


Now that we have solved for C, we can replace C in either of our two original equations with 18.  Let's use equation 1:


3A + 4C = 129 -----> 3A + 4(18) = 129 -----> 3A + 72 = 129


Next, subtract both sides by 72:


3A + 72 - 72 = 129 - 72 -----> 3A = 57


Finally, divide both sides by 3 to solve for A:


3A/3 = 57/3 -----> A = 19


Therefore, the price of an Adult ticket is $19 and the price of a Child ticket is $18.