Answer:
Step-by-step explanation:
Let x and y represent the prices of adult and student tickets, respectively. Then the revenue equations are ...
13x +9y = 138
3x +14y = 163
Solving this using a graphing calculator, we find (x, y) = (3, 11).
The price of an adult ticket is $3; the price of a student ticket is $11.
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Algebraic solution
When the numbers are "messy," as they are here, I like to use the "cross-multiplication method" for solution. I find it easier to use the order 1, x, y, rather than the order x, y, 1 as shown in the second attachment. That way, the coefficients are in their original order, with the first coefficient repeated. Note the right-side constant must be subtracted to get the form ax+by+c = 0.
[tex]\begin{array}{cccc}13&9&-138&13\\3&14&-163&3\end{array}[/tex]
Then we have ...
1/(13·14 -3·9) = x/(9·(-163) -14·(-138)) = y/((-138)·3 -(-163)·13)
x = 465/155 = 3 . . . . price of an adult ticket
y = 1705/155 = 11 . . . price of a student ticket
