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students performed in a play on a friday and a saturday. for both perfo…

Question

students performed in a play on a friday and a saturday. for both performances, adult tickets cost a dollars each and student tickets cost s dollars each.
on friday, they sold 125 adult tickets and 65 student tickets, and collected $1,200. on saturday they sold 140 adult tickets and 50 student tickets, and collected $1,230.
this situation is represented by this system of equations: { 125a + 65s = 1,200 140a + 50s = 1,230
a. what could the equation 265a + 115s = 2,430 mean in this situation?
b. the solution to the original system is the pair a = 7 and s = 5. explain why it makes sense that this pair of values is also a solution to the equation 265a + 115s = 2,430

Explanation:

Step1: Analyze the equation

The left - hand side of the equation \(265a + 115s\) is the sum of the number of adult tickets sold on Friday (\(125\)) and Saturday (\(140\)) multiplied by the cost per adult ticket (\(a\)), and the sum of the number of student tickets sold on Friday (\(65\)) and Saturday (\(50\)) multiplied by the cost per student ticket (\(s\)). The right - hand side \(2430\) is the sum of the money collected on Friday (\(1200\)) and Saturday (\(1230\)).

Step2: Interpret the meaning

The equation \(265a + 115s=2430\) means the total money collected from selling \(265\) adult tickets ( \(125 + 140\) ) and \(115\) student tickets (\(65 + 50\)) at the prices of \(a\) dollars per adult ticket and \(s\) dollars per student ticket.

Step3: Check the solution

If \(a = 7\) and \(s = 5\), substitute these values into the left - hand side of the equation \(265a+115s\):

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Since \(a = 7\) and \(s = 5\) satisfy the equation \(265a + 115s = 2430\) (because when we substitute the values of \(a\) and \(s\) into the left - hand side of the equation, we get the right - hand side value), it makes sense that \((a = 7,s = 5)\) is a solution to \(265a + 115s=2430\).

Answer:

a. The equation \(265a + 115s = 2430\) means the total money collected from selling \(265\) adult tickets ( \(125+140\) ) and \(115\) student tickets (\(65 + 50\)) at the prices of \(a\) dollars per adult ticket and \(s\) dollars per student ticket.
b. When \(a = 7\) and \(s = 5\), substituting into \(265a + 115s\) gives \(265\times7+115\times5=1855 + 575=2430\), so \((a = 7,s = 5)\) is a solution to \(265a + 115s=2430\)