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The problem provided is:

"Eight times the difference of twice a number and 10 yields the number increased by 40. Find the number."

Let's break this down into an equation step by step:

1. **Let the unknown number be represented as** \( x \).
2. **Twice the number** is \( 2x \).
3. **The difference of twice the number and 10** is \( 2x - 10 \).
4. **Eight times this difference** is \( 8(2x - 10) \).
5. This is said to **equal the number increased by 40**, which is \( x + 40 \).

Now, we can set up the equation:

\[
8(2x - 10) = x + 40
\]

### Solving the equation:

1. Distribute the 8 on the left side:
\[
16x - 80 = x + 40
\]

2. Move \( x \) to the left side by subtracting \( x \) from both sides:
\[
16x - x - 80 = 40
\]
\[
15x - 80 = 40
\]

3. Add 80 to both sides:
\[
15x = 120
\]

4. Solve for \( x \) by dividing both sides by 15:
\[
x = \frac{120}{15} = 8
\]

Thus, the number is \( \boxed{8} \).

---

Would you like further details on this solution or have any questions? Here are some related questions you might consider:

1. How do we check if the solution \( x = 8 \) is correct?
2. How does the concept of translating word problems into equations work?
3. What if the problem had a different multiplier instead of 8?
4. How would the equation change if it said "decreased by 40" instead of "increased by 40"?
5. What are other examples of linear equations involving a single variable?

**Tip:** When solving word problems, always define the variable clearly and build the equation step-by-step based on the given conditions.

Eight times the difference of twice a number and 10 yields the number increased by 40. Find the number.

Learn how to solve a linear equation where eight times the difference of twice a number and 10 equals the number increased by 40. This step-by-step solution covers key algebraic concepts and linear equation solving techniques. Suitable for students in Grades 7-9.

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