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Australia
Year 4

PRACTICE: Addition and subtraction

Lesson

Practice

Before you practice the work from this chapter, take some time to go over how we were able to use place value columns to set up a vertical algorithm. This is useful to solve  addition  and  subtraction  problems, instead of using:

  • place value models

  • number lines

  • partitioning to break our numbers up

We also looked at how to use a vertical algorithm to solve  addition  or  subtraction  where we needed to regroup our numbers.

We also saw a mix of  addition and subtraction problems of different lengths  , where we may, or may not, need to do some regrouping.

Examples

Example 1

Find the value of 150 + 250.

Worked Solution
Create a strategy

Use the vertical algorithm method.

Apply the idea

Write it in a vertical algorithm.\begin{array}{c} & &1 &5 &0 \\ &+ &2 &5 &0 \\ \hline & \\ \hline \end{array}

Add the ones column first.\begin{array}{c} & &1 &5 &0 \\ &+ &2 &5 &0 \\ \hline & & & &0 \\ \end{array}

Add the tens column. 5 + 5 = 10 . Bring down the 0 and carry the 1 to the hundreds column.\begin{array}{c} & &\text{ }^1 1 &5 &0 \\ &+ &2 &5 &0 \\ \hline & & &0 &0 \\ \end{array}

Add the hundreds column.\begin{array}{c} & &\text{ }^1 1 &5 &0 \\ &+ &2 &5 &0 \\ \hline & &4 &0 &0 \\ \end{array}

So 150 + 250 = 400 .

Example 2

Find the value of 6754 - 939.

Worked Solution
Create a strategy

Use the vertical algorithm method.

Apply the idea

Write it in a vertical algorithm.\begin{array}{c} & &6 &7 &5 &4 \\ &- & &9 &3 &9 \\ \hline & \\ \hline \end{array}

Subtract the ones column first. In the ones column we can see that 4 is less than 9, so we need to trade 1 ten from the tens place.

So we get 14-9=5 in the ones column and 5 tens becomes 4 tens in the first row.\begin{array}{c} & &6 &7 &4 &\text{ }^1 4 \\ &- & &9 &3 &9 \\ \hline & & & & &5 \\ \hline \end{array}

So for the tens column we have 4 - 3 = 1:\begin{array}{c} & &6 &7 &4 &\text{ }^1 4 \\ &- & &9 &3 &9 \\ \hline & & & &1 &5 \\ \hline \end{array}

In the hundreds column, we can see that 7 is less than 9, so we need to trade 1 thousand from the thousands place.

So we get 17-9=8 in the hundreds column and 6 thousands becomes 5 thousands in the first row.\begin{array}{c} & &5 &\text{ }^1 7 &4 &\text{ }^1 4 \\ &- & &9 &3 &9 \\ \hline & & &8 &1 &5 \\ \hline \end{array}

And for the thousands place we just bring down the 5:\begin{array}{c} & &5 &\text{ }^1 7 &4 &\text{ }^1 4 \\ &- & &9 &3 &9 \\ \hline & &5 &8 &1 &5 \\ \hline \end{array}

So 6754 - 939 = 5815.

Example 3

Find the value of 7215 + 7039.

Worked Solution
Create a strategy

Use the vertical algorithm method.

Apply the idea

Write it in a vertical algorithm.\begin{array}{c} & & &7 &2 &1 &5 \\ &+ & &7 &0 &3 &9 \\ \hline & \\ \hline \end{array}

Add the ones column first. 5 + 9 =14. Bring down the 4 and carry the 1 to the tens column.

\begin{array}{c} & & &7 &2 &1 \text{ }^1 &5 \\ &+ & &7 &0 &3 &9 \\ \hline & & & & & &4 \\ \hline \end{array}

Then add the tens column, 1 + 1 + 3 = 5:\begin{array}{c} & & &7 &2 &1 \text{ }^1 &5 \\ &+ & &7 &0 &3 &9 \\ \hline & & & & &5 &4 \\ \hline \end{array}

Then add the hundreds column, 2 + 0 = 2:\begin{array}{c} & & &7 &2 &1 \text{ }^1 &5 \\ &+ & &7 &0 &3 &9 \\ \hline & & & &2 &5 &4 \\ \hline \end{array}

Then add the thousands column, 7 + 7 =14. Bring down the 4 and carry the 1 to the ten thousands column.

\begin{array}{c} & &\text{ }^1 &7 &2 &1 \text{ }^1 &5 \\ &+ & &7 &0 &3 &9 \\ \hline & & &4 &2 &5 &4 \\ \hline \end{array}

Then add the ten thousands column, 1 + 0 = 1:\begin{array}{c} & &\text{ }^1 &7 &2 &1 \text{ }^1 &5 \\ &+ & &7 &0 &3 &9 \\ \hline & &1 &4 &2 &5 &4 \\ \hline \end{array}

So 7215 + 7039 = 14\,254 .

Idea summary

When we use a vertical algorithm, we need to make sure our digits are lined up so they are in the correct place. If we need to regroup, we regroup to the left for addition, and to the right for subtraction.

Outcomes

AC9M4N06

develop efficient strategies and use appropriate digital tools for solving problems involving addition and subtraction, and multiplication and division where there is no remainder

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