The sum of two numbers is 39. One number is 2 times as large as the other. What are the numbers?

Let's solve this problem step by step.

Let the two numbers be \( x \) and \( y \). According to the problem:

1. The sum of the two numbers is 39:
   \[
   x + y = 39
   \]

2. One number is 2 times as large as the other:
   \[
   x = 2y
   \]

Now, substitute the expression for \( x \) from the second equation into the first equation:

\[
2y + y = 39
\]

Combine like terms:

\[
3y = 39
\]

Solve for \( y \):

\[
y = \frac{39}{3} = 13
\]

Now that we have \( y = 13 \), substitute this value back into the equation \( x = 2y \):

\[
x = 2 \times 13 = 26
\]

So, the two numbers are \( x = 26 \) and \( y = 13 \).

Would you like more details or have any questions?

Here are 8 related questions to explore:

1. What if the sum of the two numbers was 50 instead of 39?
2. How would the solution change if one number was 3 times as large as the other?
3. Can you solve for the two numbers if their difference is given instead of their sum?
4. What if the relationship was \( x = \frac{1}{2}y \) instead of \( x = 2y \)?
5. How do you solve this problem using a system of equations?
6. What are the properties of the numbers if the sum is a negative number?
7. How does the problem change if both numbers are negative?
8. Can you solve the problem using graphical methods?

**Tip:** When dealing with problems involving multiple conditions, try expressing one variable in terms of the other to simplify the problem.

Learn how to solve the problem where the sum of two numbers is 39 and one number is 2 times the other. Step-by-step solution provided, covering key algebraic concepts such as linear equations. Suitable for students in Grades 7-9.

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