One number is 2/5 of another number the sum of the two numbers is 35. Find the two numbers use a, to separate your answers.

Let's denote the two numbers as \( x \) and \( y \). According to the problem, we have the following conditions:

1. \( x = \frac{2}{5}y \)  (One number is \(\frac{2}{5}\) of the other)
2. \( x + y = 35 \)  (The sum of the two numbers is 35)

We can substitute the first equation into the second equation:

\[
\frac{2}{5}y + y = 35
\]

Next, combine the terms on the left-hand side:

\[
\frac{2}{5}y + \frac{5}{5}y = 35
\]

\[
\frac{7}{5}y = 35
\]

To solve for \( y \), multiply both sides by \(\frac{5}{7}\):

\[
y = 35 \times \frac{5}{7} = 25
\]

Now that we have \( y \), substitute it back into the first equation to find \( x \):

\[
x = \frac{2}{5} \times 25 = 10
\]

So, the two numbers are **10** and **25**.

**Answer**: \( 10 \) **,** \( 25 \).

Would you like more details on any part of this solution? Here are five related questions:

1. How do you find the least common multiple (LCM) of two numbers?
2. How would you solve if the relationship was \( \frac{3}{4} \) instead of \( \frac{2}{5} \)?
3. How can you verify the solution by substitution?
4. What happens if the sum of the two numbers was different, say 50?
5. How would you solve it if the problem involved decimals instead of fractions?

**Tip:** When dealing with fractions, it's helpful to multiply both sides by the denominator to simplify calculations.

Discover how to solve a problem involving fractional relationships where one number is 2/5 of another, and their sum is 35. This step-by-step guide covers algebraic techniques suitable for students in Grades 6-8, including solving linear equations and handling fractional expressions.

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