Three adults and four children must pay $95. Two adults and three children must pay $67. Find the price of the tickets for adult and child.

**a = price of adult**

c = price of child

c = price of child

3a + 4c = 95 {three adults and 4 children is $95}

2a + 3c = 67 {two adults and 3 children is $67}

-6a - 8c = -190 {multiplied top equation by -2}

__6a + 9c = 201__{multiplied bottom equation by 3}

c = 11 {added the two equations together}

3a + 4c = 95 {top equation}

3a + 4(11) = 95 {substituted 11, in for c, into top equation}

3a + 44 = 95 {multiplied 4 by 11}

3a = 51 {subtracted 44 from each side}

a = 17 {divided each side by 3}

**adult ticket is $17**

child ticket is $11

child ticket is $11

Also, the graph of the systems of equations verifies

the point of intersection (the solution) to be at (17,11)

*- Algebra House*