## Algebra: A Combined Approach (4th Edition)

Let $a$ be the price of an adult's ticket. Let $c$ be the price of a child's ticket. Three adults and four children must pay 159 dollars. $3a+4c=159$ Two adults and three children must pay 112 dollars. $2a+3c=112$ $3a+4c=159$ $2a+3c=112$ Multiply first equation by 2 and second equation by -3. $6a+8c=318$ $-6a-9c=-336$ Add both equations. $-c=-18$ $c=18$ Solve for $a$ $2a+3c=112$ $2a+3(18)=112$ $2a+54=112$ $2a=58$ $a=29$ An adult's ticket is 29 dollars. A child's ticket is 18 dollars.