Let's solve this step by step:

Let the original two-digit number be represented as $10a + b$, where:

• $a$ is the tens digit,
• $b$ is the ones digit.

Step 1: Write the given information as equations.

• The sum of the digits is 7: $a + b = 7$

• When the digits are reversed, the new number is 9 less than the original number: $10b + a = (10a + b) - 9$ Simplify this equation: $10b + a = 10a + b - 9$ Subtract $b$ and $a$ from both sides: $9b = 9a - 9$ Divide through by 9: $b = a - 1$

Step 2: Solve the system of equations.

You now have the system:

1. $a + b = 7$
2. $b = a - 1$

Substitute $b = a - 1$ into the first equation: $a + (a - 1) = 7$ Simplify: $2a - 1 = 7$ Add 1 to both sides: $2a = 8$ Divide by 2: $a = 4$

Now substitute $a = 4$ back into $b = a - 1$: $b = 4 - 1 = 3$

Step 3: Find the original number.

The original number is: $10a + b = 10(4) + 3 = 43$

Final Answer:

The original number is 43.

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Would you like a more detailed explanation, or do you have any questions about the steps?

Here are five related questions you might find interesting:

1. How can we check the correctness of the solution?
2. What would happen if the sum of the digits were a different number, say 8?
3. How can you generalize this process to solve similar problems with other conditions?
4. Can you solve for the digits of a three-digit number under similar constraints?
5. How would the equation change if the new number were 5 less than the original number?

Tip: When solving problems like this, it's useful to represent the number as $10a + b$ and use the relationships between the digits to form equations.