The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is th digit number?

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:
\[
\begin{aligned}
1.\quad & a + b = 15 \\
2.\quad & a - b = 3
\end{aligned}
\]

### 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.

Learn how to solve a two-digit number problem where the sum of the digits is 15 and the difference between the digits is 3. Follow a step-by-step solution method using algebraic equations. Suitable for students in Grades 7-9.

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