Let's denote the two-digit number as $10a + b$, where $a$ is the tens digit and $b$ is the units digit.

We are given the following conditions:

1. The digit in the unit's place is four times the digit in the tens place: $b = 4a$.
2. The sum of the digits is 10: $a + b = 10$.

Step 1: Substitute the first condition into the second equation

From $b = 4a$, substitute into the sum equation $a + b = 10$: $a + 4a = 10$ $5a = 10$ $a = \frac{10}{5} = 2$

Step 2: Find $b$ using $b = 4a$

$b = 4 \times 2 = 8$

Step 3: Form the two-digit number

The digits are $a = 2$ and $b = 8$, so the two-digit number is $28$.

The two-digit number is 28.

Would you like more details or have any questions?

Here are some related questions:

1. What if the sum of the digits was different?
2. How would the solution change if the digit in the unit's place was three times the digit in the tens place?
3. Can you find a two-digit number where the digits have a specific ratio?
4. What if the units digit was less than the tens digit by a certain amount?
5. How can we generalize this problem for any ratio and sum?

Tip: When solving problems with digit relationships, substitution is a powerful tool to simplify the equations and find the solution quickly.