To solve this, we can set up a system of linear equations. Let:

• $x$ be the price of a senior citizen ticket
• $y$ be the price of a child ticket

From the problem, we know the following two conditions:

1. On the first day, 3 senior citizen tickets and 1 child ticket were sold for a total of P38. This gives the equation: $3x + y = 38$

2. On the second day, 3 senior citizen tickets and 2 child tickets were sold for a total of P52. This gives the equation: $3x + 2y = 52$

We now have a system of two equations:

3x + y &= 38 \quad \text{(Equation 1)} \\ 3x + 2y &= 52 \quad \text{(Equation 2)} \end{aligned}$$ ### Step 1: Eliminate $$3x$$ by subtracting Equation 1 from Equation 2. $$(3x + 2y) - (3x + y) = 52 - 38$$ Simplifying this: $$y = 14$$ ### Step 2: Substitute $$y = 14$$ into Equation 1. $$3x + 14 = 38$$ Subtract 14 from both sides: $$3x = 24$$ Now, solve for $$x$$: $$x = \frac{24}{3} = 8$$ ### Conclusion: - The price of a senior citizen ticket ($$x$$) is P8. - The price of a child ticket ($$y$$) is P14. Would you like any further explanation on the solution, or have more questions on similar problems? Here are 5 related questions for practice: 1. If the school sold 4 senior citizen tickets and 3 child tickets on a third day for P64, what would the price of the tickets be? 2. How would the solution change if the second day's total was P58 instead of P52? 3. Solve a system of equations where the total revenue on day one is P40 and day two is P60 with different ticket prices. 4. If the price of senior citizen tickets increased by P2, how would the prices of the tickets change? 5. If the total sales on day 1 were split 50/50 between senior citizens and children, how would this impact the price calculation? **Tip:** When solving systems of equations, always look for opportunities to eliminate one variable by adding or subtracting the equations. This can make the process much quicker!