Answer:
It is not 37, it is 34
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.