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​​ALGEBRA eXAMPLES

Price per ticket word problem

12/6/2014

 
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)


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