To solve the problem, we need to establish two equations based on the information provided.

Let's denote:

• $x$ as the number of children.
• $y$ as the number of apples.

Step 1: Formulate the first equation

The first scenario states that each apple is cut into 4 equal pieces. Each child receives 1 piece, leaving 3 pieces.

So, the total number of pieces obtained from $y$ apples is $4y$. If each of the $x$ children receives 1 piece, the equation will be: $x = 4y - 3$

Step 2: Formulate the second equation

The second scenario states that each apple is cut into 3 equal pieces. Two pieces were not enough for all the children.

So, the total number of pieces obtained from $y$ apples is $3y$. Since two pieces were not enough, it implies that the total number of children is two less than the total number of pieces: $x = 3y + 2$

Step 3: Set up the simultaneous equations

The two equations we derived are: $x = 4y - 3 \quad \text{(Equation 1)}$ $x = 3y + 2 \quad \text{(Equation 2)}$

Step 4: Solve the simultaneous equations

Set the equations equal to each other to eliminate $x$: $4y - 3 = 3y + 2$ Subtract $3y$ from both sides: $y - 3 = 2$ Add 3 to both sides: $y = 5$

Now that we have $y$, substitute it into one of the original equations to find $x$. Using Equation 2: $x = 3(5) + 2 = 15 + 2 = 17$

Final Answer:

• The number of children ($x$) is 17.
• The number of apples ($y$) is 5.

Would you like more details or have any questions?

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Here are 5 related questions that might interest you:

1. What happens if each apple is cut into 5 equal pieces?
2. How would the equations change if two pieces were left in the first scenario?
3. Can you solve a similar problem where each apple is divided into 6 pieces?
4. What is the solution if instead of apples, oranges are divided into 4 and 3 pieces?
5. How would the solution differ if the number of leftover pieces in the first scenario was 4 instead of 3?

Tip: Always double-check your equations by substituting your final values back into the original conditions to ensure accuracy.