4.06 Subtraction with regrouping

Use the subtraction algorithm.

Write the subtraction in a vertical algorithm.\begin{array}{c} & &9 &5 \\ &- &5 &1 \\ \hline & \\ \hline \end{array}

Subtract the smallest place value first. So 5 - 1 = 4.\begin{array}{c} & &9 &5 \\ &- &5 &1 \\ \hline & & &4 \\ \hline \end{array}

Then subtract the next place value, 9- 5 = 4.\begin{array}{c} & &9 &5 \\ &- &5 &1 \\ \hline & &4 &4 \\ \hline \end{array}

So, 95- 51 = 44.

Use the subtraction algorithm with regrouping.

Write the subtraction in a vertical algorithm.\begin{array}{c} & &5 &8 \\ &- &2 &9 \\ \hline & \\ \hline \end{array}

Begin with the ones column. Since we don't have enough ones to subtract 9 from 8, we need to trade 1 ten from the tens place. \begin{array}{c} & &4 & \text{}^1 8 \\ &- &2 &9 \\ \hline & & \\ \hline \end{array}

Now we can subtract 9 from 18 in the ones column and 5 tens becomes 4 tens.\begin{array}{c} & &4 & \text{}^1 8 \\ &- &2 &9 \\ \hline & & & 9 \\ \hline \end{array}

Then subtract the numbers in the tens column, 4- 2 = 2.\begin{array}{c} & &4 & \text{}^1 8 \\ &- &2 &9 \\ \hline & &2 & 9 \\ \hline \end{array}

So, 58 - 29=29.