Let's define the two-digit number as $10a + b$, where:

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

Step 1: Set up the equations

We are given two pieces of information:

1. The sum of the digits is 11: $a + b = 11$

2. When the digits are reversed, the new number is 5 greater than 3 times the original number: The number with reversed digits is $10b + a$. The condition is: $10b + a = 3(10a + b) + 5$

Step 2: Solve the system of equations

From the first equation:

$a + b = 11 \quad \text{(Equation 1)}$ We can express $b$ as: $b = 11 - a$

Substitute $b = 11 - a$ into the second equation:

$10b + a = 3(10a + b) + 5$ Substitute $b = 11 - a$ into this: $10(11 - a) + a = 3(10a + (11 - a)) + 5$ Simplifying both sides: $110 - 10a + a = 3(10a + 11 - a) + 5$ $110 - 9a = 3(9a + 11) + 5$ Distribute on the right-hand side: $110 - 9a = 27a + 33 + 5$ $110 - 9a = 27a + 38$ Now, move all terms involving $a$ to one side: $110 - 38 = 27a + 9a$ $72 = 36a$ Solving for $a$: $a = \frac{72}{36} = 2$

Step 3: Find $b$

From Equation 1, $a + b = 11$, so: $2 + b = 11$ $b = 9$

Step 4: The original number

The original number is $10a + b = 10(2) + 9 = 29$.

Final Answer:

The original number is 29.

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Would you like further details or clarifications? Here are some related questions to explore:

1. How would the problem change if the sum of the digits was different?
2. What happens if the reversed number is 4 greater than 3 times the original number?
3. How would you solve this if it were a three-digit number instead?
4. What is the importance of setting up a system of equations in this type of problem?
5. How can you check if the solution is correct?

Tip: Always label the digits of the number carefully (e.g., tens digit as $a$ and ones digit as $b$) to set up a clear system of equations.