3-Digit Subtraction with Borrowing worksheet - Free Printable

The task involves solving subtraction problems using the place value system (hundreds, tens, and ones). Let's solve each problem step by step.

Problem 1:

\[
\begin{array}{c@{}c@{}c}
& 9 & 2 & 5 \\
- & 3 & 8 & 9 \\
\hline
\end{array}
\]

#### Step-by-step solution:
1. Ones place: \(5 - 9\)
- Since 5 is less than 9, we need to borrow 1 from the tens place.
- After borrowing, we have \(15 - 9 = 6\).
- The tens place now becomes \(2 - 1 = 1\).

2. Tens place: \(1 - 8\)
- Since 1 is less than 8, we need to borrow 1 from the hundreds place.
- After borrowing, we have \(11 - 8 = 3\).
- The hundreds place now becomes \(9 - 1 = 8\).

3. Hundreds place: \(8 - 3 = 5\).

So, the result is:
\[
\boxed{536}
\]

Problem 2:

\[
\begin{array}{c@{}c@{}c}
& 8 & 4 & 2 \\
- & 4 & 7 & 8 \\
\hline
\end{array}
\]

#### Step-by-step solution:
1. Ones place: \(2 - 8\)
- Since 2 is less than 8, we need to borrow 1 from the tens place.
- After borrowing, we have \(12 - 8 = 4\).
- The tens place now becomes \(4 - 1 = 3\).

2. Tens place: \(3 - 7\)
- Since 3 is less than 7, we need to borrow 1 from the hundreds place.
- After borrowing, we have \(13 - 7 = 6\).
- The hundreds place now becomes \(8 - 1 = 7\).

3. Hundreds place: \(7 - 4 = 3\).

So, the result is:
\[
\boxed{364}
\]

Problem 3:

\[
\begin{array}{c@{}c@{}c}
& 5 & 2 & 3 \\
- & 2 & 9 & 7 \\
\hline
\end{array}
\]

#### Step-by-step solution:
1. Ones place: \(3 - 7\)
- Since 3 is less than 7, we need to borrow 1 from the tens place.
- After borrowing, we have \(13 - 7 = 6\).
- The tens place now becomes \(2 - 1 = 1\).

2. Tens place: \(1 - 9\)
- Since 1 is less than 9, we need to borrow 1 from the hundreds place.
- After borrowing, we have \(11 - 9 = 2\).
- The hundreds place now becomes \(5 - 1 = 4\).

3. Hundreds place: \(4 - 2 = 2\).

So, the result is:
\[
\boxed{226}
\]

Problem 4:

\[
\begin{array}{c@{}c@{}c}
& 6 & 3 & 4 \\
- & 3 & 7 & 9 \\
\hline
\end{array}
\]

#### Step-by-step solution:
1. Ones place: \(4 - 9\)
- Since 4 is less than 9, we need to borrow 1 from the tens place.
- After borrowing, we have \(14 - 9 = 5\).
- The tens place now becomes \(3 - 1 = 2\).

2. Tens place: \(2 - 7\)
- Since 2 is less than 7, we need to borrow 1 from the hundreds place.
- After borrowing, we have \(12 - 7 = 5\).
- The hundreds place now becomes \(6 - 1 = 5\).

3. Hundreds place: \(5 - 3 = 2\).

So, the result is:
\[
\boxed{255}
\]

Problem 5:

\[
\begin{array}{c@{}c@{}c}
& 8 & 3 & 2 \\
- & 5 & 5 & 6 \\
\hline
\end{array}
\]

#### Step-by-step solution:
1. Ones place: \(2 - 6\)
- Since 2 is less than 6, we need to borrow 1 from the tens place.
- After borrowing, we have \(12 - 6 = 6\).
- The tens place now becomes \(3 - 1 = 2\).

2. Tens place: \(2 - 5\)
- Since 2 is less than 5, we need to borrow 1 from the hundreds place.
- After borrowing, we have \(12 - 5 = 7\).
- The hundreds place now becomes \(8 - 1 = 7\).

3. Hundreds place: \(7 - 5 = 2\).

So, the result is:
\[
\boxed{276}
\]

Problem 6:

\[
\begin{array}{c@{}c@{}c}
& 7 & 2 & 6 \\
- & 4 & 4 & 7 \\
\hline
\end{array}
\]

#### Step-by-step solution:
1. Ones place: \(6 - 7\)
- Since 6 is less than 7, we need to borrow 1 from the tens place.
- After borrowing, we have \(16 - 7 = 9\).
- The tens place now becomes \(2 - 1 = 1\).

2. Tens place: \(1 - 4\)
- Since 1 is less than 4, we need to borrow 1 from the hundreds place.
- After borrowing, we have \(11 - 4 = 7\).
- The hundreds place now becomes \(7 - 1 = 6\).

3. Hundreds place: \(6 - 4 = 2\).

So, the result is:
\[
\boxed{279}
\]

Final Answers:

\[
\boxed{536, 364, 226, 255, 276, 279}
\]