[tex]\begin{gathered} \text{Given} \\ a+c=65 \\ 7a+4c=335 \end{gathered}[/tex]
Solve by substitution, use the first equation and solve in terms of c
[tex]\begin{gathered} a+c=65 \\ c=65-a \end{gathered}[/tex]
Substitute this to the second equation and solve for a
[tex]\begin{gathered} 7a+4c=335 \\ 7a+4(65-a)=335 \\ 7a+260-4a=335 \\ 7a-4a=335-260 \\ 3a=75 \\ \frac{3a}{3}=\frac{75}{3} \\ a=25 \end{gathered}[/tex]
Substitute the value of a back to the first equation
[tex]\begin{gathered} a+c=65 \\ 25+c=65 \\ c=65-25 \\ c=40 \end{gathered}[/tex]
With a = 25, and c = 40, we can conclude that there are 25 adult tickets, and 40 child tickets sold on Saturday.
