Question 1160648
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From the condition, you have these two equations


    3A + 4C = 111    (1)

    2A + 3C =  79    (2)


To solve this system, I will apply the Elimination method.

For it, I multiply equation (1) by 2  (both sides), and multiply equation (2) by 3 (both sides).

My goal is to make coefficients at "A" equal.

You will get then


    6A + 8C = 222    (3)

    6A + 9C = 237    (4)


Now subtract equation (3) from equation (4).  The terms "6A" will cancel each other,
and you will get single equation for only one unknown "C"


         9C - 8C = 237 - 222

          C       = 15.


So, we just made half of the job: we found the price of the child ticket, C = 15.


Now substitute this value of "C" into equation (1) to get "A"


    3A + 4*15 = 111,

which gives you

    3A = 111 - 5*15 = 111 - 60 = 51.


Hence,  A = 51/3 = 17.


<U>ANSWER</U>.  The child ticket price is $15;  the adult ticket price is $17.
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Solved.


You may check the solution by substituting the found values into the original equations.