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 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 Also, the graph of the systems of equations verifies the point of intersection (the solution) to be at (17,11)  Algebra House Comments are closed.

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