Subtracting 1-Digit from 2-Digit Numbers - Second Grade Mathematics

Subtracting a single-digit number from a two-digit number expands your mental subtraction skills, allowing you to count backward, decompose numbers, and cross decade boundaries with ease. Depending on the numbers involved, you will encounter two situations: subtracting without crossing a decade (such as 48 - 5), where the tens place remains untouched, and subtracting across a decade (such as 42 - 5), where you must break open a ten to find your answer. In this chapter, you will master both cases using mental decomposition, open number lines, and base-ten models.

Case 1: Subtracting Without Crossing a Decade

When the ones digit of the two-digit number is greater than or equal to the single digit you are subtracting, the tens digit never changes.

Solving 57 - 4:
Tens in 57: 5 tens (value = 50)
Ones in 57: 7 ones
Subtract:   4 ones

Subtract Ones:  7 - 4 = 3
Recombine:      50 + 3 = 53

Simple Ones Subtraction

Because 7 ones is larger than 4 ones, you have plenty of loose unit cubes to take from. You simply subtract 7 - 4 = 3, and reattach the 5 tens: 50 + 3 = 53. More examples: 39 - 6: Subtract ones 9 - 6 = 3. Attach to 30 = 33. 88 - 5: Subtract ones 8 - 5 = 3. Attach to 80 = 83. 65 - 2: Subtract ones 5 - 2 = 3. Attach to 60 = 63.

Case 2: Crossing a Decade (Break Apart to Land on 10)

What happens when you need to solve 52 - 5? In 52, there are only 2 loose ones. You cannot take 5 ones from 2 ones without unbundling a ten!

Solving 52 - 5 by Landing on the Decade:

Step 1: Look at 52. How many steps back to the friendly decade 50?
        52 needs to lose 2 ones to land on 50.

Step 2: Decompose the single digit 5 into 2 and 3.
        5 breaks into 2 and 3.

Step 3: Subtract the 2 to land on 50.
        52 - 2 = 50

Step 4: Subtract the remaining 3 from 50.
        50 - 3 = 47

The Two-Hop Backward Journey

Think of crossing into the lower decade like stepping down stairs. In the problem 63 - 7: Start at 63. Step back 3 stairs to reach the level landing of 60. You had 7 stairs to descend in total, and you already stepped down 3. You have 4 more stairs to step down: 60 - 4 = 56!

Number Line Jumps for 63 - 7:
          -4               -3
    <------------+   <------------+
----+------------+---+------------+--->
   56           60               63

Base-Ten Unbundling Model

Let us look at what physically happens with base-ten blocks during 42 - 5.

Start with 42:    | | | |      . .

Problem: We need to cross out 5 dots.
Trouble: There are only 2 dots!

Action: Take 1 ten rod | and unbundle it into 10 unit dots:
Now we have:      | | |        . . . . . . . . . .   . .
                               (10 new dots)         (2 original dots)
Total Ones: 12 dots!

Cross out 5 dots: 12 - 5 = 7 dots remain.
Tens remaining:   3 lines remain (30).
Result: 30 + 7 = 37!

Why Unbundling Works

Unbundling did not change the total value of 42. It simply converted 4 tens and 2 ones into 3 tens and 12 ones. Once 12 ones were available, removing 5 ones was easy, leaving 7 ones and 3 tens = 37.

Practice Exercises

Exercise 1: Solve 78 - 6. Explain whether this problem crossed a decade boundary, and state which digit changed.

Exercise 2: Solve 43 - 7 using the decompose-to-land-on-ten strategy. Show how the number 7 was broken apart.

Exercise 3: Solve 81 - 5 using an open number line. Describe both backward jumps starting from 81.

Exercise 4: A bakery had 34 chocolate muffins in a display case. During the morning rush, 8 muffins were purchased. How many chocolate muffins remained in the case?

Solutions and Explanations

Solution to Exercise 1: To solve 78 - 6, subtract the ones: 8 - 6 = 2. Attach to 70: 70 + 2 = 72. This problem did not cross a decade boundary because the ones digit (8) was large enough to subtract 6 without needing to unbundle a ten. Only the ones digit changed (from 8 to 2), while the tens digit remained 7.

Solution to Exercise 2: To solve 43 - 7, look at 43. It needs to lose 3 to land on 40. Decompose 7 into 3 and 4. First subtract 3: 43 - 3 = 40. Next subtract the remaining 4 from 40: 40 - 4 = 36. The final answer is 36.

Solution to Exercise 3: On an open number line, start at 81. Make a first jump backward of 1 unit to land on the friendly decade 80. Since 5 breaks into 1 and 4, make a second backward jump of 4 units from 80 to land on 76. The difference is 76.

Solution to Exercise 4: To find the remaining muffins, calculate 34 - 8. Break 8 into 4 and 4. First, 34 - 4 = 30. Then, 30 - 4 = 26. There were 26 chocolate muffins remaining in the display case.