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We are tasked with separating the number 76 into three addends with specific conditions. Let's define the three addends as:

- Let the third number be \( x \).
- The first number is 15 more than the third, so the first number is \( x + 15 \).
- The second number is 31 more than the third, so the second number is \( x + 31 \).

Now, according to the problem, the sum of these three numbers equals 76:

\[
(x + 15) + (x + 31) + x = 76
\]

We can solve this equation step by step:

1. Combine like terms:
\[
3x + 46 = 76
\]

2. Subtract 46 from both sides:
\[
3x = 30
\]

3. Divide both sides by 3:
\[
x = 10
\]

Thus, the third number is \( x = 10 \).

Now, substitute \( x = 10 \) to find the other two numbers:
- The first number is \( 10 + 15 = 25 \).
- The second number is \( 10 + 31 = 41 \).

So, the three numbers are **25**, **41**, and **10**, and their sum is:
\[
25 + 41 + 10 = 76
\]

This satisfies all the conditions of the problem.

Would you like further details or explanations?

Here are five related questions for deeper understanding:
1. How would the equation change if the second addend was 40 more than the third instead of 31?
2. What if the total sum was 100 instead of 76, but the same conditions applied?
3. How can you verify that the sum of the three addends always equals 76?
4. How would you set up the equation if the first number was 10 less than the third?
5. Can you think of another scenario where we use algebra to split a number into different parts?

**Tip**: When solving word problems, always try to express the relationships between unknowns clearly through algebraic expressions. This makes solving equations much easier!

Separate the number 76 into three addends so that the first is 15 more than the third, and the second is 31 more than the third.

Learn how to separate the number 76 into three addends where the first is 15 more than the third and the second is 31 more than the third. This algebraic problem includes a step-by-step solution, focusing on setting up and solving linear equations. Suitable for students in Grades 6-8.

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