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The problem describes a situation where:

1. Laura buys 5 chocolate bars and 2 boxes of bonbons, spending R$26.00 in total.
2. Ana buys 3 chocolate bars and 4 boxes of bonbons, spending R$31.00 in total.

The goal is to find the price of each chocolate bar and each box of bonbons by solving the system of equations derived from these purchases.

Let's define:
- $ x $ as the price of each chocolate bar.
- $ y $ as the price of each box of bonbons.

### Step 1: Set up the equations
From the information provided, we have the following two equations:

1. For Laura:  
   \[
   5x + 2y = 26
   \]

2. For Ana:  
   \[
   3x + 4y = 31
   \]

### Step 2: Solve the system of equations
We can use either substitution or elimination to solve this system. Let's proceed with elimination for simplicity.

1. Multiply the first equation by 2 to align the $ y $-terms:

   \[
   10x + 4y = 52
   \]

2. Now subtract the second equation from this result:

   \[
   (10x + 4y) - (3x + 4y) = 52 - 31
   \]
   \[
   7x = 21
   \]
   \[
   x = 3
   \]

3. Substitute $ x = 3 $ into the first equation:

   \[
   5(3) + 2y = 26
   \]
   \[
   15 + 2y = 26
   \]
   \[
   2y = 11
   \]
   \[
   y = 5.5
   \]

### Solution
- Each chocolate bar ($ x $) costs **R$3.00**.
- Each box of bonbons ($ y $) costs **R$5.50**.

Would you like a detailed verification of the solution or have any questions?

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Here are 5 related questions to explore further:
1. What if Laura bought 7 chocolate bars instead of 5, how would the system change?
2. How would the solution change if Ana paid R$34.00 instead of R$31.00 for her items?
3. Can this system be solved using matrix operations? How?
4. What method would you use to solve a similar problem with three variables?
5. How can this problem be represented graphically?

**Tip:** When solving systems of equations, double-check your arithmetic, as small errors can lead to incorrect results.

Laura and Ana went to a candy store. Laura bought 5 chocolate bars and 2 boxes of bonbons, spending R$26.00 in total. Ana bought 3 chocolate bars and 4 boxes of bonbons, spending R$31.00 in total. The chocolate bars and bonbons both had the same price. Find the price of each chocolate bar and each box of bonbons by solving the following system of equations.

Learn how to solve a system of equations involving chocolate bars and bonbons bought by Laura and Ana. This problem involves setting up and solving a system of linear equations using the elimination method. Perfect for students in Grades 9-10 who are studying algebra and systems of equations.

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