The drama club is selling tickets to their play to raise money for the show's expenses.
Each student ticket sells for $7.50 and each adult ticket sells for $12.50. The
auditorium can hold at most 96 people. The drama club must make no less than $990
from ticket sales to cover the show's costs. If 59 adult tickets were sold, determine the
minimum number of student tickets that the drama club must sell in order to meet
the show's expenses. If there are no possible solutions, submit an empty answer.

It is not 37, it is 34

Step-by-step explanation:

\underline{\text{Define Variables:}}

Define Variables:

May choose any letters.

\text{Let }s=

Let s=

\,\,\text{the number of student tickets sold}

the number of student tickets sold

\text{Let }a=

Let a=

\,\,\text{the number of adult tickets sold}

the number of adult tickets sold

\text{\textquotedblleft at most 96 people"}\rightarrow \text{96 or fewer tickets}

“at most 96 people"→96 or fewer tickets

Use a \le≤ symbol

Therefore the total number of tickets sold, s+as+a, must be less than or equal to 96:96:

s+a\le 96

s+a≤96

\text{\textquotedblleft no less than \$990"}\rightarrow \text{\$990 or more}

“no less than $990"→$990 or more

Use a \ge≥ symbol

Each student ticket sells for $7.50, so ss student tickets will bring in 7.50s7.50s dollars. Each adult ticket sells for $12.50, so aa adult tickets will bring in 12.50a12.50a dollars. Therefore, the total amount of revenue 7.50s+12.50a7.50s+12.50a must be greater than or equal to \$990:$990:

7.50s+12.50a\ge 990

7.50s+12.50a≥990

\text{Plug in }59\text{ for }a\text{ and solve each inequality:}

Plug in 59 for a and solve each inequality:

Colton worked 59 adult tickets

\begin{aligned}s+a\le 96\hspace{10px}\text{and}\hspace{10px}&7.50s+12.50a\ge 990 \\ s+\color{green}{59}\le 96\hspace{10px}\text{and}\hspace{10px}&7.50s+12.50\left(\color{green}{59}\right)\ge 990 \\ s\le 37\hspace{10px}\text{and}\hspace{10px}&7.50s+737.50\ge 990 \\ \hspace{10px}&7.50s\ge 252.50 \\ \hspace{10px}&s\ge 33.67 \\ \end{aligned}

s+a≤96and

s+59≤96and

s≤37and

7.50s+12.50a≥990

7.50s+12.50(59)≥990

7.50s+737.50≥990

7.50s≥252.50

s≥33.67

\text{The values of }s\text{ that make BOTH inequalities true are:}

The values of s that make BOTH inequalities true are:

\{34,\ 35,\ 36,\ 37\}

{34, 35, 36, 37}

Therefore the minimum number of student tickets that the drama club must sell is 34.

5

step-by-step explanation: