The given problem provides the equation $3x + 5y = 500$, where:

• $x$ is the number of child tickets sold (each costing $3),
• $y$ is the number of adult tickets sold (each costing $5), and
• The school aims to raise exactly $500 from ticket sales.

1. Solving the equation for $x$:

To solve for $x$, we need to isolate $x$ in the equation:

$3x + 5y = 500$

Subtract $5y$ from both sides:

$3x = 500 - 5y$

Now, divide both sides by 3 to solve for $x$:

$x = \frac{500 - 5y}{3}$

This expression gives the number of child tickets sold in terms of the number of adult tickets $y$.

2. When it might be helpful to rewrite the equation this way:

Rewriting the equation for $x$ allows us to express the number of child tickets sold based on the number of adult tickets sold. This could be helpful in scenarios where the number of adult tickets is known or easier to estimate, and we need to determine how many child tickets need to be sold to meet the fundraising goal of $500. This form is useful when solving for one variable while keeping the other as a parameter.

Would you like more details on any specific part of this solution, or have any further questions? Here are five related questions that could extend the discussion:

1. How would you solve the equation for $y$ instead of $x$?
2. What happens if the number of adult tickets is fixed at a specific value, like 50? How many child tickets would be needed?
3. Can you graph the equation $3x + 5y = 500$ and explain what the intercepts represent?
4. How would this equation change if child tickets were sold for $4 instead of $3?
5. What does it mean if $x$ or $y$ is negative in the context of this problem?

Tip: When solving equations with two variables, isolating one variable can help simplify calculations when specific values for the other variable are known.