// Numbas version: exam_results_page_options {"name": "Addition and Subtraction", "metadata": {"description": "

Addition and Subtraction by hand questions, whole numbers and decimals. Fully worked solutions in feedback.

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Addition calculations to be carried out by hand, full worked solutions in feedback.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Addition of numbers of any size.

Calculate the following without using a calculator.

", "advice": "

Addition of numbers

Part a)

To add numbers we need to take into account their place value, so line numbers up placing the hundreds, tens and units in vertical columns.

\\[\\begin{align} \\var{a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{1cm}\\end{align}\\]

Addition is started in the right hand column, the units. $\\var{au}+\\var{bu}=\\var{au+bu}$. Any tens in the answer to this are then carried across to be added to the tens total. In this case the answer has $\\var{((au+bu)-mod(au+bu,10))/10}$ in the tens column to carry across and $\\var{mod((au+bu),10)}$ in the units column which can go straight in for the answer.

\\[\\begin{align} \\var{a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\ _\\var{((au+bu)-mod(au+bu,10))/10} \\var{mod(au+bu,10)}&\\\\\\end{align}\\]

The next column is the tens column, $\\var{at}+\\var{bt}=\\var{at+bt}$ Add to this anything that was carried over from the units, $\\var{((au+bu)-(mod(au+bu,10)))/10}$, makes it $\\var{at+bt+((au+bu)-mod(au+bu,10))/10}$. As we are adding the tens column, the $\\var{mod(at+bt+((au+bu)-mod(au+bu,10))/10,10)}$ is placed in the answer in this position and the $\\var{(at+bt+((au+bu)-mod(au+bu,10))/10-(mod(at+bt+((au+bu)-mod(au+bu,10))/10,10)))/10}$ is carried across to the hundreds column.

\\[\\begin{align} \\var{a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\ _\\var{(at+bt-mod(at+bt,10))/10} \\var{mod(at+bt+((au+bu)-mod(au+bu,10))/10,10)}\\var{mod(au+bu,10)}&\\\\\\end{align}\\]

Then finally the hundreds column. There is only one value in this column so all that is needed is for any digits carried over from the total of the tens column need to be added to this value. So this is $\\var{bh}+\\var{(at+bt-mod(at+bt,10))/10}=\\var{bh+(at+bt-mod(at+bt,10))/10}$. This is then placed in the answer in the appropriate position, in the hundreds column, if the answer has two digits then the first of these is carried across and placed in the thousands position.

\\[\\begin{align} \\var{a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\var{a+b}\\end{align}\\]

If there were more values then the process would continue until you had added each column in turn.

Part b)

\\[\\begin{align} \\var{2*a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{1cm}\\end{align}\\]

Starting with the units $\\var{mod(2*a,10)}+\\var{bu}=\\var{mod(2*a,10)+bu}$.

So $\\var{((mod(2*a,10)+bu)-mod(mod(2*a,10)+bu,10))/10}$ is carried across to the tens column and $\\var{mod((mod(2*a,10)+bu),10)}$ in the units column answer.

\\[\\begin{align} \\var{2*a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\ _\\var{((mod(2*a,10)+bu)-mod(mod(2*a,10)+bu,10))/10} \\var{mod(mod(2*a,10)+bu,10)}&\\\\\\end{align}\\]

The next column, $\\var{(mod(2*a,100)-mod(2*a,10))/10}+\\var{bt}=\\var{(mod(2*a,100)-mod(2*a,10))/10+bt}$

Add to this anything that was carried over from the units, $\\var{(mod(2*a,100)-mod(2*a,10))/10+bt}+\\var{((mod(2*a,10)+bu)-(mod(mod(2*a,10)+bu,10)))/10}=\\var{(mod(2*a,100)-mod(2*a,10))/10+bt+((mod(2*a,10)+bu)-mod(mod(2*a,10)+bu,10))/10}$.

So $\\var{mod((mod(2*a,100)-mod(2*a,10))/10+bt+((mod(2*a,10)+bu)-mod(mod(2*a,10)+bu,10))/10,10)}$ is placed in the answer and $\\var{((mod(2*a,100)-mod(2*a,10))/10+bt-mod((mod(2*a,100)-mod(2*a,10))/10+bt,10))/10}$ is carried across to the hundreds column.

\\[\\begin{align} \\var{2*a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\ _\\var{((mod(2*a,100)-mod(2*a,10))/10+bt-mod((mod(2*a,100)-mod(2*a,10))/10+bt,10))/10} \\var{mod((mod(2*a,100)-mod(2*a,10))/10+bt+((mod(2*a,10)+bu)-mod(mod(2*a,10)+bu,10))/10,10)}\\var{mod(mod(2*a,10)+bu,10)}&\\\\\\end{align}\\]

Similarly for the hundreds column, $\\var{(mod(2*a,1000)-mod(2*a,100))/100}+\\var{bh}+\\var{((mod(2*a,100)-mod(2*a,10))/10+bt-mod((mod(2*a,100)-mod(2*a,10))/10+bt,10))/10}=\\var{(mod(2*a,1000)-mod(2*a,100))/100+bh+((mod(2*a,100)-mod(2*a,10))/10+bt-mod((mod(2*a,100)-mod(2*a,10))/10+bt,10))/10}$ This is then placed in the answer in the appropriate position, in the hundreds column, if the answer has two digits then the first of these is carried across and placed in the thousands position.

\\[\\begin{align} \\var{2*a}&\\\\\\underline{+\\hspace{0.5cm}\\var{b}}&\\\\=\\hspace{0.5cm}\\var{2*a+b}\\end{align}\\]

Addition of decimal numbers

Part c)

The process for addition of decimals is the same as that for whole numbers, lining up place values and starting addition with the right hand column. The important thing to remember when adding decimals is to ensure that all the numbers are aligned by the decimal point, this ensures each column contains the right place value.

Whole numbers are lined up so that they are directly to the left of the point. For ease you can add 0's in spaces where there are no numbers, making each number have the same number of decimal places.

\\[\\begin{align} \\var{a3}&\\\\\\var{c3}&\\\\\\underline{+\\hspace{0.5cm}\\var{d}}&\\\\=\\hspace{2cm}\\end{align}\\]

Then carrying out addition down each column, starting with right hand column $0+0+\\var{sigformat(mod((1000*d),10),1)}=\\var{sigformat(mod((1000*d),10),1)}$.

\\[\\begin{align} \\var{a3}&\\\\\\var{c3}&\\\\\\underline{+\\hspace{0.5cm}\\var{d}}&\\\\=\\hspace{2cm}\\var{sigformat(mod((1000*d),10),1)}&\\end{align}\\]

Continuing in this way $0+\\var{mod((100*c),10)}+\\var{(mod((1000*d)-mod(1000*d,10),100))/10}=\\var{mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10}$

In this case the answer has $\\var{(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10}$ to carry across and $\\var{mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)}$ which can go straight in for the answer.

\\[\\begin{align} \\var{a3}&\\\\\\var{c3}&\\\\\\underline{+\\hspace{0.5cm}\\var{d}}&\\\\=\\hspace{2cm}\\ _\\var{sigformat((mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10,1)} \\var{sigformat(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10),1)}\\var{sigformat(mod(1000*d,10),1)}&\\\\\\end{align}\\]

The next column is $0+\\var{(mod(100*c,100)-mod(100*c,10))/10}+\\var{(mod(1000*d,1000)-mod(1000*d,100))/100}=\\var{(mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100}$

Adding anything carried over $\\var{(mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100}+\\var{(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10}=\\var{(mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10}$

\\[\\begin{align} \\var{a3}&\\\\\\var{c3}&\\\\\\underline{+\\hspace{0.5cm}\\var{d}}&\\\\=\\hspace{2cm}\\ _\\var{((mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10-mod((mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10,10))/10} .\\var{mod((mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10,10)}\\var{mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)}\\var{sigformat(mod(1000*d,10),1)}&\\\\\\end{align}\\]

So for the above the carry across is $\\var{((mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10-mod((mod(100*c,100)-mod(100*c,10))/10+(mod(1000*d,1000)-mod(1000*d,100))/100+(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10-(mod(mod((100*c),10)+(mod((1000*d)-mod((1000*d),10),100))/10,10)))/10,10))/10}$ and the addition continues in the same way as for the whole numbers to produce

\\[\\begin{align} \\var{a3}&\\\\\\var{c3}&\\\\\\underline{+\\hspace{0.5cm}\\var{d}}&\\\\=\\hspace{2cm}\\var{d+a+c}&\\end{align}\\]

Part d)

Again the first thing to do is to line the numbers up by thier decimal points. You do not need to put the numbers into size order but some people find this helpful.

\\[\\begin{align} \\var{j}&\\\\\\var{k}&\\\\\\underline{+\\hspace{0.5cm}\\var{l}}&\\\\=\\hspace{2cm}\\end{align}\\]

Then carrying out addition down each column, starting with right hand column $0+0+\\var{sigformat(mod((1000*l),10),1)}=\\var{sigformat(mod((1000*l),10),1)}$.

\\[\\begin{align} \\var{j}&\\\\\\var{k}&\\\\\\underline{+\\hspace{0.5cm}\\var{l}}&\\\\=\\hspace{2cm}\\var{sigformat(mod((1000*l),10),1)}&\\end{align}\\]

Continuing in this way $0+\\var{mod((100*k1),10)}+\\var{(mod((1000*l)-mod(1000*l,10),100))/10}=\\var{mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10}$

In this case the answer has $\\var{(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10}$ to carry across and $\\var{mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)}$ which can go straight in for the answer.

\\[\\begin{align} \\var{j}&\\\\\\var{k}&\\\\\\underline{+\\hspace{0.5cm}\\var{l}}&\\\\=\\hspace{2cm}\\ _\\var{(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10} \\var{mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)}\\var{sigformat(mod(1000*l,10),1)}&\\\\\\end{align}\\]

The next column is $0+\\var{(mod(100*k1,100)-mod(100*k1,10))/10}+\\var{(mod(1000*l,1000)-mod(1000*l,100))/100}=\\var{(mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100}$

Adding anything carried over $\\var{(mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100}+\\var{(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10}=\\var{(mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100+(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10}$

\\[\\begin{align} \\var{j}&\\\\\\var{k}&\\\\\\underline{+\\hspace{0.5cm}\\var{l}}&\\\\=\\hspace{2cm}\\ _\\var{((mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100+(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10-mod((mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100+(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10,10))/10} .\\var{mod((mod(100*k1,100)-mod(100*k1,10))/10+(mod(1000*l,1000)-mod(1000*l,100))/100+(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10-(mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)))/10,10)}\\var{mod(mod((100*k1),10)+(mod((1000*l)-mod((1000*l),10),100))/10,10)}\\var{sigformat(mod(1000*l,10),1)}&\\\\\\end{align}\\]

And carrying on in the same way until all columns have been completed to give

\\[\\begin{align} \\var{j}&\\\\\\var{k}&\\\\\\underline{+\\hspace{0.5cm}\\var{l}}&\\\\=\\hspace{2cm}\\var{j1+k1+l}&\\end{align}\\]

Part e)

Again lining the numbers up around the decimal point and adding each column in turn gives

\\[\\begin{align} \\var{b3}&\\\\\\var{l}&\\\\\\underline{+\\hspace{0.5cm}\\var{c3}}&\\\\=\\hspace{2cm}\\var{b+l+c}&\\end{align}\\]

Additional resources

For further examples and questions

Multi Digit Addition - Khan Academy

Adding Decimals - Khan Academy

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$\\var{a}+\\var{b}$=[[0]]

$\\var{b}+\\var{a*2}$ =[[0]]

$\\var{a}+\\var{c}+\\var{d}$=[[0]]

$\\var{j1}+\\var{k1}+\\var{l}$=[[0]]

$\\var{b}+\\var{l}+\\var{c}$=[[0]]

Subtraction questions, whole and decimal, to be carried out by hand. Full worked solutions given in feedback.

", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

Subtraction.

Work out the following without using a calculator.

", "advice": "

Subtraction by hand

To carry out subtraction by hand the easiest way is to line the numbers up one above the other. When lining them up make sure that each place value is lined up correctly. Remember that the first number in the subtraction is placed above the second. The order of subtraction is important and it must be remembered throughout the calculation that it is the top number minus the bottom number.

Part a)

i)

Lining these up gives

\\[\\begin{align}\\var{a}&\\\\\\underline{-{\\hspace 0.5cm}\\var{b}}&\\\\\\end{align}\\]

Carrying out subtraction down each column in turn starting with the furthest right and placing the answer of these two digits underneath each column gives

$\\var{z}-\\var{v}=\\var{z-v}$,

\\[\\begin{align}\\var{a}&\\\\\\underline{-{\\hspace 0.5cm}\\var{b}}&\\\\\\var{z-v}&\\\\\\end{align}\\]

And for the other columns $\\var{y}-\\var{u}=\\var{y-u}$ and $\\var{x}-\\var{t}=\\var{x-t}$

\\[\\begin{align}\\var{a}&\\\\\\underline{-{\\hspace 0.5cm}\\var{b}}&\\\\\\var{a-b}&\\\\\\end{align}\\]

ii)

Lining these up gives

\\[\\begin{align}\\var{c}&\\\\\\underline{-{\\hspace 0.5cm}\\var{d}}&\\\\\\end{align}\\]

Carrying out subtraction down each column in turn starting with the furthest right and placing the answer of these two digits underneath each column gives

$\\var{y}-\\var{u}=\\var{y-u}$,

\\[\\begin{align}\\var{c}&\\\\\\underline{-{\\hspace 0.5cm}\\var{d}}&\\\\\\var{y-u}&\\\\\\end{align}\\]

And for the other columns $\\var{z}-\\var{v}=\\var{z-v}$ and $\\var{x}-\\var{t}=\\var{x-t}$

\\[\\begin{align}\\var{c}&\\\\\\underline{-{\\hspace 0.5cm}\\var{d}}&\\\\\\var{c-d}&\\\\\\end{align}\\]

Part b)

In part a for each column the number at the top was equal to or greater than the number below it. In this case the subtraction did not need any rearranging or consideration of surrounding columns. This is not often the case with subtraction and when the top number of a column is less than the number beneath it some rearranging may be required. At this point remembering that the column is part of a bigger calculation enables the digits to be changed so that the top number is bigger in each column. It can be helpful to think of it in terms of money and changing a larger domination coin into smaller ones, 10p = 10 x 1p. If the digit at the top of a column is less than the digit below it then this can be changed by looking to the column to the left. If this is not 0 at the top you can change one from this column into 10 in your column.

i)

Lining up the numbers in the same way gives

\\[\\begin{align}\\var{f}&\\\\\\underline{-{\\hspace 0.5cm}\\var{g}}\\\\\\end{align}\\]

Looking at the furthest right column the calculation we want to perform is $\\var{t}-\\var{z}$, but the number at the top is smaller than the number at the bottom. We need to change this by reducing the number at the top in the column to the left of it by 1 and adding 10 to the top of this column. So the top row now becomes $\\var{y},\\var{x-1},\\var{t+10}$ and the subtraction is performed on the following numbers.

\\[\\begin{align}\\var{y}^{\\;\\var{x-1}}\\rlap{/}\\var{x}^{\\;1}\\var{t}&\\\\\\underline{-{\\hspace 0.5cm}\\var{t}\\ \\;\\;\\var{u}\\ \\;\\; \\var{z}}&\\\\\\end{align}\\]

So now carrying out the subtraction down each column gives

$\\var{t+10}-\\var{z}=\\var{t+10-z}$, $\\var{x-1}-\\var{u}=\\var{x-1-u}$, $\\var{y}-\\var{t}=\\var{y-t}$

\\[\\begin{align}\\var{y}^{\\;\\var{x-1}}\\rlap{/}\\var{x}^{\\;1}\\var{t}&\\\\\\underline{-{\\hspace 0.5cm}\\var{t}\\ \\;\\;\\var{u}\\ \\;\\;\\var{z}}&\\\\\\var{y-t}\\ \\;\\;\\var{x-1-u}\\ \\;\\;\\var{t+10-z}&\\\\\\end{align}\\]

ii)

Again lining up the numbers gives

\\[\\begin{align}\\var{h}&\\\\\\underline{-{\\hspace 0.5cm}\\var{l}}\\\\\\end{align}\\]

Looking at each column and exchanging along the top where the bottom number is larger the subtraction becomes

\\[\\begin{align}\\var{x}^{\\;\\var{u-1}}\\rlap{/}\\var{u}^{\\;1}\\var{v}&\\\\\\underline{-{\\hspace 1cm}\\var{y}\\ \\;\\; \\var{z}}&\\\\\\end{align}\\]

There is still a column with the top number smaller than the bottom, so the process is repeated, carried out in all positions from right to left where the top number is smaller than the bottom.

\\[\\begin{align}^{\\;\\var{x-1}}\\rlap{/}\\var{x}^{\\;\\var{u-1+10}}\\rlap{/}\\var{u}^{\\;1}\\var{v}&\\\\\\underline{-{\\hspace 1cm}\\var{y}\\ \\;\\; \\var{z}}&\\\\\\end{align}\\]

Carrying out subtraction down each column in turn starting with the furthest right and placing the answer of these two digits underneath each column gives

\\[\\begin{align}^{\\;\\var{x-1}}\\rlap{/}\\var{x}^{\\;\\var{u-1+10}}\\rlap{/}\\var{u}^{\\;1}\\var{v}&\\\\\\underline{-{\\hspace 1cm}\\var{y}\\ \\; \\var{z}}&\\\\\\var{x-1}\\ \\;\\;\\; \\var{u-1+10-y}\\ \\; \\var{v+10-z}&\\\\\\end{align}\\]

Part c)

With decimals the key point is to line up the numbers by the decimal point and it can be made easier by adding 0's in to any gaps.

i)

So lining up the numbers in this case gives

\\[\\begin{align}\\var{x}\\var{y}.\\var{z}0&\\\\\\underline{-{\\hspace 0.5cm}\\var{t}.\\var{u}\\var{v}}&\\\\\\end{align}\\]

The same method is used as before, looking at each column in turn and if the top number is smaller than the bottom then change the top numbers by taking one from the column to the left and adding 10 to the current column so the top number becomes more than the bottom in that column.

\\[\\begin{align}\\var{x}\\ \\;\\; \\var{y}.^{\\var{z-1}}\\rlap{/}\\var{z}^{\\ \\;1}0&\\\\\\underline{-{\\hspace 0.5cm}\\var{t}.\\ \\; \\var{u}\\ \\;\\; \\var{v}}&\\\\\\end{align}\\]

Carrying out the calculation the answer is found. Make sure that the decimal point is placed in the correct place in the answer

\\[\\begin{align}\\var{x}\\ \\;\\; \\var{y}.^{\\var{z-1}}\\rlap{/}\\var{z}^{\\ \\;1}0&\\\\\\underline{-{\\hspace 0.5cm}\\var{t}.\\ \\; \\var{u}\\ \\;\\; \\var{v}}&\\\\\\var{x}\\ \\;\\; \\var{y-t}.\\ \\; \\var{z-1-u}\\ \\;\\; \\var{10-v}&\\\\\\end{align}\\]

ii)

\\[\\begin{align}\\var{t}\\var{u}.\\var{z}0&\\\\\\underline{-{\\hspace 0.5cm}\\var{y}.\\var{x}\\var{t}}&\\\\\\end{align}\\]

\\[\\begin{align}\\var{t}\\ \\;\\; \\var{u}.^{\\var{z-1}}\\rlap{/}\\var{z}^{\\ \\;1}0&\\\\\\underline{-{\\hspace 0.5cm}\\var{y}.\\ \\; \\var{x}\\ \\;\\; \\var{t}}&\\\\\\end{align}\\]

Carying out the calculation the answer is found . Make sure that the decimal point is placed in the correct place in the answer.

\\[\\begin{align}\\var{t}\\ \\;\\;^{\\var{u-1}} \\rlap{/}\\var{u}.^{\\var{z-1+10}}\\rlap{/}\\var{z}^{\\ \\;1}0&\\\\\\underline{-{\\hspace 0.5cm}\\var{y}.\\;\\ \\; \\var{x}\\ \\;\\; \\var{t}}&\\\\\\end{align}\\]

\\[\\begin{align}^{\\var{t-1}}\\rlap{/}\\var{t}\\ \\;\\;^{\\var{u-1+10}} \\rlap{/}\\var{u}.^{\\var{z-1+10}}\\rlap{/}\\var{z}^{\\ \\;1}0&\\\\\\underline{-{\\hspace 0.5cm}\\var{y}.\\;\\ \\; \\var{x}\\ \\;\\; \\var{t}}&\\\\\\end{align}\\]

Now carrying out the calculations in each column gives

\\[\\begin{align}^{\\var{t-1}}\\rlap{/}\\var{t}\\ \\;\\;^{\\var{1}} \\var{u}.^{\\var{z-1}}\\rlap{/}\\var{z}^{\\ \\;1}0&\\\\\\underline{-{\\hspace 0.5cm}\\var{y}.\\ \\; \\var{x}\\ \\;\\; \\var{t}}&\\\\\\end{align}\\]

Now carrying out the calculations in each column gives

Additional resources

For further examples and questions

Multi Digit Subtraction - Khan Academy

Subtracting Decimals - Khan Academy

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