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

• $a$ is the tens digit,
• $b$ is the units (or ones) digit.

From the problem, we have two conditions:

1. Sum of the digits is 14: $a + b = 14$

2. Twice the tens digit, increased by two, equals four times the units digit: $2a + 2 = 4b$

Step-by-Step Solution

Let's solve this system of equations.

1. From the first equation, solve for $b$: $b = 14 - a$

2. Substitute $b = 14 - a$ into the second equation: $2a + 2 = 4(14 - a)$ Expanding and simplifying: $2a + 2 = 56 - 4a$ $6a = 54$ $a = 9$

3. Substitute $a = 9$ back into the equation $b = 14 - a$: $b = 14 - 9 = 5$

Conclusion

The two-digit number is: $10a + b = 10 \times 9 + 5 = 95$

Answer

The number is 95.

Would you like a more detailed explanation or have any other questions?

---

Here are 5 related questions you might explore:

1. How can we verify the solution by plugging the values back into the original conditions?
2. What if the sum of the digits were different—how would that affect the solution?
3. How could this problem be solved using different methods, such as trial and error?
4. Can a similar approach work for three-digit numbers with specific conditions?
5. How can we generalize this approach to find a number given conditions on digit sums?

Tip: In word problems, defining variables and translating conditions into equations is key to setting up the solution.