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 two conditions:

1. The sum of the digits is 15: $a + b = 15$.
2. The difference between the digits is 3: $a - b = 3$.

Step 1: Solve the system of equations

We have the following system of equations:

1.\quad & a + b = 15 \\ 2.\quad & a - b = 3 \end{aligned}$$ ### Step 2: Add the two equations to eliminate $$b$$ $$(a + b) + (a - b) = 15 + 3$$ $$2a = 18$$ $$a = \frac{18}{2} = 9$$ ### Step 3: Substitute $$a = 9$$ into the first equation to find $$b$$ $$9 + b = 15$$ $$b = 15 - 9 = 6$$ ### Step 4: Form the two-digit number The digits are $$a = 9$$ and $$b = 6$$, so the two-digit number is $$96$$. **The two-digit number is 96.** 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 difference between the digits was 5 instead of 3? 3. Can you find a two-digit number where the digits multiply to a given number? 4. What are other two-digit numbers with the same digit sum but a different difference? 5. How can we express any two-digit number algebraically? **Tip:** When solving systems of equations, try to eliminate one variable by adding or subtracting the equations. This simplifies finding the solution.