We are given two equations relating the number of adult tickets and children's tickets to the total cost. We need to find the price of an adult ticket and a child's ticket. Equation 1: Three adults and four children pay $101. Equation 2: Two adults and three children pay $72.

1. Problem Description

We are given two equations relating the number of adult tickets and children's tickets to the total cost. We need to find the price of an adult ticket and a child's ticket.

Equation 1: Three adults and four children pay $

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1. Equation 2: Two adults and three children pay $

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2. Solution Steps

Let $a$ be the price of an adult ticket and $c$ be the price of a child's ticket.

We can write the given information as a system of two linear equations:

$3a + 4c = 101$

$2a + 3c = 72$

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 to eliminate $a$.

$2(3a + 4c) = 2(101)$

$3(2a + 3c) = 3(72)$

This gives us:

$6a + 8c = 202$

$6a + 9c = 216$

Subtract the first equation from the second equation:

$(6a + 9c) - (6a + 8c) = 216 - 202$

$c = 14$

Now that we have the price of a child's ticket, $c=14$, substitute this into either equation to find the price of an adult's ticket. Let's use the second equation:

$2a + 3c = 72$

$2a + 3(14) = 72$

$2a + 42 = 72$

$2a = 72 - 42$

$2a = 30$

$a = 15$

The price of an adult ticket is $15 and the price of a child's ticket is$

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3. Final Answer

The price of a child's ticket is $14.