Answer :
SOLUTION
We will start by drafting out the 3 equations needed in solving this question.
Let matinee ticket be x, student ticket be y, and regular ticket be z.
A matinee admission cost $5 each; student admission cost S6 each, and regular admission cost S8 each and everything summing up to $32100 can be expressed mathematically as:
[tex]5x+6y+8z=\text{ 32100}\ldots\text{.eqn 1}[/tex]Twice as many student tickets were sold as matinee tickets sold can be expressed mathematically as:
[tex]y=2x\ldots eqn\text{ 2}[/tex]And the total number of tickets can be expressed mathematically as:
[tex]x+y+z=4800\ldots\text{eqn 3}[/tex]Now we will solve all three equations to know how many of each type of ticket was sold.
Substituting eqn 2 into eqn 1 and eqn 3 we will have:
[tex]\begin{gathered} \text{eqn 1 becomes:} \\ 5x+6(2x)+8z=32100 \\ 5x+12x+8z=32100 \\ 17x+8z=32100\ldots\text{.eqn 4} \\ \\ \text{eqn 2 becomes} \\ x+2x+z=4800 \\ 3x+z=4800\ldots\text{eqn 5} \end{gathered}[/tex]Now we can solve eqn 4 and 5 simultaneously to get the value of x and z
[tex]\begin{gathered} 17x+8z=32100\ldots\text{eqn 4} \\ 3x+z=4800\ldots\text{eqn 5} \\ z=4800-3x \\ \\ \text{eqn 4 becomes:} \\ 17x+8(4800-3x)=32100 \\ 17x+38400-24x=32100 \\ 38400-7x=32100 \\ 38400-32100=7x \\ 6300=7x \\ \frac{6300}{7}=\frac{7x}{7x} \\ 900=x \end{gathered}[/tex]So the number of matinee tickets sold is 900.
[tex]\begin{gathered} z=4800-3x \\ z=4800-3(900)_{} \\ z=4800-2700 \\ z=2100 \\ \\ y=2x \\ y=2(900) \\ y=1800 \end{gathered}[/tex]The number of student tickets sold is 1800 and the number of regular tickets sold is 2100.
