The problem describes a scenario where Ann Marie Jones is pricing train fares for a group trip to New York. We are given two pieces of information: - Three adults and four children must pay $104. - Two adults and three children must pay $74. We are asked to find the price of an adult's ticket and the price of a child's ticket.

1. Problem Description

The problem describes a scenario where Ann Marie Jones is pricing train fares for a group trip to New York. We are given two pieces of information:

- Three adults and four children must pay $

1

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4. - Two adults and three children must pay $

7

4. We are asked to find the price of an adult's ticket and the price of a child's ticket.

2. Solution Steps

Let $a$ be the price of an adult's ticket and $c$ be the price of a child's ticket. We can set up a system of two linear equations with two variables:

$3a + 4c = 104$

$2a + 3c = 74$

We can solve this system of equations using substitution or elimination. Let's use elimination. Multiply the first equation by 2 and the second equation by 3:

$2(3a + 4c) = 2(104)$

$3(2a + 3c) = 3(74)$

This simplifies to:

$6a + 8c = 208$

$6a + 9c = 222$

Now, subtract the first equation from the second equation:

$(6a + 9c) - (6a + 8c) = 222 - 208$

$c = 14$

Now that we know the price of a child's ticket, we can substitute $c = 14$ into either of the original equations to solve for $a$. Let's use the first equation:

$3a + 4(14) = 104$

$3a + 56 = 104$

$3a = 104 - 56$

$3a = 48$

$a = \frac{48}{3}$

$a = 16$

3. Final Answer

The price of a child's ticket is $

1

4. The price of an adult's ticket is $16.