Use the fact that there are 5 tickets in total for one equation, and that the total cost is $52 for another. One equation will involve just $a$ and $c$, and the other will involve 12, 8, $a$, and $c$.

From the simpler equation (the one with just $a$ and $c$), solve for one variable in terms of the other and substitute into the money equation. Then simplify to find the value of the remaining variable.

In Desmos, type a + c = 5 on one line and 12a + 8c = 52 on the next line. Desmos will interpret these as two lines in the $a$–$c$ plane.

Look for the point where the two lines intersect. Click on the intersection; Desmos will show its coordinates $(a, c)$. The second coordinate (the $c$-value) is the number of child tickets.

Let $a$ be the number of adult tickets and $c$ be the number of child tickets that Dominic bought.

Use the two facts in the problem:

• Total number of tickets is 5:

• Adult tickets cost $12 and child tickets cost $8, for a total of $52:

Now you have a system of two equations with two variables.

Solve the system using substitution (you could also use elimination):

From $a + c = 5$, solve for $a$:

Substitute this into the money equation:

Distribute and simplify:

Subtract 60 from both sides:

Divide both sides by $-4$ to find $c$.

From $-4c = -8$, dividing by $-4$ gives $c = 2$.

So Dominic bought 2 child tickets.