the difference of two numbers is 12. two fifths of the greater number is six more than one third of the lesser number. find both number

We are given two conditions for the numbers:

1. **The difference of two numbers is 12.**
   \[
   x - y = 12
   \]
   where \( x \) is the greater number and \( y \) is the lesser number.

2. **Two-fifths of the greater number is six more than one-third of the lesser number.**
   \[
   \frac{2}{5}x = \frac{1}{3}y + 6
   \]

Now, we have two equations:
1. \( x - y = 12 \)
2. \( \frac{2}{5}x = \frac{1}{3}y + 6 \)

### Solving the system of equations:
From equation (1), we can express \( x \) in terms of \( y \):
\[
x = y + 12
\]

Substitute this into equation (2):
\[
\frac{2}{5}(y + 12) = \frac{1}{3}y + 6
\]

Now, expand and simplify:
\[
\frac{2}{5}y + \frac{24}{5} = \frac{1}{3}y + 6
\]

Multiply through by 15 to eliminate the denominators:
\[
15\left(\frac{2}{5}y + \frac{24}{5}\right) = 15\left(\frac{1}{3}y + 6\right)
\]
This gives:
\[
6y + 72 = 5y + 90
\]

Simplifying further:
\[
6y - 5y = 90 - 72
\]
\[
y = 18
\]

Now substitute \( y = 18 \) into \( x = y + 12 \):
\[
x = 18 + 12 = 30
\]

### Final Answer:
The two numbers are **30** and **18**.

Would you like further details or explanations?

Here are 5 related questions:
1. How can we solve the same system of equations graphically?
2. What happens if the difference between the numbers changes?
3. Can we solve for \( x \) and \( y \) using substitution or elimination in all systems of equations?
4. How does multiplying both sides of the equation by the least common denominator help?
5. Can these equations be solved using matrix methods?

**Tip:** Always check your solution by substituting values back into the original equations!

The difference of two numbers is 12. Two fifths of the greater number is six more than one third of the lesser number. Find both numbers.

Learn how to solve a system of linear equations where the difference of two numbers is 12, and two-fifths of the greater number is six more than one-third of the lesser number. This problem involves key algebraic concepts and uses substitution to find both numbers. Suitable for students in Grades 8-10.

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