Let's solve this problem step by step.

Given:

1. The sum of the digits of a two-digit number is 6.
2. When the digits are reversed, the number decreases by 18.
3. Find the original number.

Let the two-digit number be represented as:

$10x + y$ where:

• $x$ is the tens digit,
• $y$ is the units digit.

Step 1: Formulate the equations

1. The sum of the digits is 6: $x + y = 6$

2. When the digits are reversed, the number decreases by 18: The reversed number is $10y + x$, so: $(10x + y) - (10y + x) = 18$ Simplifying: $9x - 9y = 18 \quad \text{or} \quad x - y = 2$

Step 2: Solve the equations

We now have a system of two equations:

1. $x + y = 6$,
2. $x - y = 2$.

Add the two equations: $(x + y) + (x - y) = 6 + 2$ $2x = 8 \quad \Rightarrow \quad x = 4$

Substitute $x = 4$ into $x + y = 6$: $4 + y = 6 \quad \Rightarrow \quad y = 2$

Step 3: Find the original number

The original number is: $10x + y = 10(4) + 2 = 42$

Final Answer:

The original number is 42.

Would you like a detailed explanation of any part of this?
Here are 5 related questions to expand your understanding:

1. How do we interpret the reversal of digits mathematically for multi-digit numbers?
2. Can you generalize this approach to find any two-digit number with given properties?
3. What if the reversed number increases instead of decreasing? How would the equations change?
4. How do you check your solution for accuracy in this type of problem?
5. Could we solve this graphically or using matrices for more complex cases?

Tip: Always double-check your system of equations for consistency when solving word problems!