Let the number of children attended be c;
Let the number of adults attended be a;
and let the number of students that attended be s;
Then;
![\begin{gathered} c+s+a=305\ldots\ldots\ldots\ldots\text{equation 1} \\ 5c+7s+12a=2212\ldots\ldots.equation\text{ 2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vzqul5vt3vavr5iloyp8urikcvmn207hd9.png)
But there are half as many adults as there are children, then;
![\begin{gathered} a=(1)/(2)c \\ c=2a\ldots\ldots\ldots\ldots\ldots\text{. equation 3} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/of1b5bjmqm2mg5f2v1z4d90ht1lxorf329.png)
By substituting equation 3 in equation 1 and equation 2, we have;
![\begin{gathered} 2a+s+a=305 \\ 3a+s=305\ldots\ldots\ldots\ldots\text{ equation 4} \\ 5(2a)+7s+12a=2212 \\ 22a+7s=2212\ldots\ldots\ldots\text{.}\mathrm{}\text{equation 5} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mp7dfa65avgegoexeggbnqz61bdfm8p3of.png)
Solving equation 4 and equation 5 simultaneously, we have;
![\begin{gathered} \text{From equation 4;} \\ s=305-3a \\ \text{put s=305-3a in equation 5;} \\ 22a+7(305-3a)=2212 \\ a=2212-2135 \\ a=77 \\ \text{Put a=77 in s=305-3a;} \\ s=305-3(77) \\ s=74 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/9yuihtan4bfx1jbqlzhgl3sq9a7jqrvmdt.png)
Then, we substitute the value of a and s in equation 1 to get the value of c;
![\begin{gathered} c+74+77=305 \\ c=154 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/5r5rqa02k3hgard6w6jdbyiey9407c8mr4.png)
154 children attended
74 students attended
77 adults attended.