Question 930461
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}
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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
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Also, the graph of the two equations shows the point of intersection as (17,11)
*[illustration Graph_of_Equations].