Adding decimals is almost as easy as adding whole numbers.  We need to pay extra attention to place values, and we must never write extra zeros, which is a waste of time.

Let us try 6.7 + 5.3.  Regrouping, we have (6 + 5) ones plus (7 + 3) tenths.  Simplifying, we have 11 ones + 10 tenths, or 12 ones.  The answer is 12.

Now let us try 1.6 + 2.1 + 12.3.  Regrouping, we have 1 ten, plus (1 + 2 + 2) ones, plus (6 + 1 + 3) tenths.  Simplifying, we have 1 ten + 5 ones + 10 tenths, or 10 + 5 + 1.  The answer is 16.

Let us try 17.23 + 1.18 + 10.39.  Working from right to left, we have (3 + 8 + 9) = 20 hundredths.  We do not write the zero in the hundredths place.  We regroup the 2 with the tenths.

So (2 + 1 + 3) plus the regrouped 2 tenths = 8 tenths.  We write an 8 with a decimal to its left.

Then we have (7 + 1 + 0) ones, or 8 ones.  We write the 8 to the left of the decimal.

Then we have (1 + 1) tens, or 2 tens.  We write the 2 at the left.  The answer is 28.8.

When asked to subtract whole numbers, we again must recognize and group together the place values of each number.

Let us try 6457 – 312.  This example does not require any regrouping.  Working from right to left, we have (7 – 2) ones, so we write a 5.

We have (5 – 1) tens, so we write a 4 to the left of the 5.  We have (4 – 3) hundreds, so we write a 1 to the left of the 4.

We have 6 thousands, so we write the 6 on the left.  The answer is 6145.

Let us try an example with regrouping, such as 562 – 275.  We must think of this as (5 – 2 hundreds), plus (6 – 7 tens), plus (2 – 5).

Regrouping, we get (4 – 2) hundreds, plus (16 – 7) tens, plus (2 – 5).  Regrouping again, we get 2 hundreds, plus (15 – 7) tens, plus (12 – 5).

Our answer is 2 hundred, 8 tens, and 7, or 287.

When asked to add a group of whole numbers, we must recognize and group together the place values of each number.

Let us try 129 + 198 + 321.  The solution is (1 + 1 + 3) hundreds, plus (2 + 9 + 2) tens, plus (9 + 8 + 1) ones.

This is simplified as 5 hundreds plus 13 tens plus 18 ones.  13 tens is equal to 1 hundred plus 3 tens.  18 ones is equal to 1 ten plus 8 ones.

The answer is 5 hundreds + 1 hundred, plus 3 tens + 1 ten, plus 8 ones.  6 hundreds plus 4 tens plus 8 ones is simply 648.

Let us try an example where each number has different lengths, such as 2413 + 22 + 917.

Working from right to left, we have (7 + 1 + 3) ones, or 12 ones.  We write the 2 and regroup the 1 with the tens.  Then we have (1 + 2 + 1) tens, plus the regrouped 1, = 5 tens.

We write the 5 to the left of the 2.  Then we have (9 + 4) hundreds, or 13 hundreds. 

We write the 3 to the left of the 5 and regroup the 1 with the thousands. 

We have 2 plus the regrouped 1, or 3 thousands, so we write the 3 on the left.  The answer is 3352.

Sometimes, when given an addition problem, it is easier to solve it by regrouping the numbers by place values, and then adding and subtracting.

For example, the number 999 can also be thought of as (1000 – 1).  4997 can be thought of as (5000 – 3).  2994 can be thought of as (3000 – 6).

If we are given the problem 999 + 4997 + 2994, it is easier to regroup each number than to add each digit, one by one.

Think of 999 + 4997 + 2994 as (1000 – 1) + (5000 – 3) + (3000 – 6).  Or (1000 + 5000 + 3000) – 1 – 3 – 6.  This is the same as (1000 + 5000 + 3000) – (1 + 3 + 6).

We regroup this as 9000 – 10, and the answer is 8990.

Let us try another example.  What is 96 + 898 + 697?  We regroup this as (100 – 4) + (900 – 2) + (700 – 3).

This is the same as 1700 – 9, so the answer is 1691.

Building on the ideas from Round and PlaceValue, let us learn about expanded notation.  Numbers written in expanded notation are grouped by ones, tens, thousands, and so on.

Parentheses in standard notation are helpful, but not required.  The number 1974 can be thought of or written as (1 x 1000) + (9 x 100) + (7 x 10) + (4 x 1).

Or 1974 can be written as 1 x 1000 + 9 x 100 + 7 x 10 + 4 x 1.  The digits are grouped with (or multiplied by) their place value first, and then added together.

Let us try an example.  What is (4 x 1000) + (7 x 100) + (9 x 10) + (1 x 1)? 

The answer is 4 thousand, 7 hundred, ninety, and one.  We write this as 4791.

We must not be confused if the place values are given to us out of order.  For example, 4791 could also be written as (9 x 10) + (4 x 1000) + (1 x 1) + (7 x 100).

Expanded notation can also apply to decimals.  Try (3 x 1) + (1 x 0.1) + (4 x 0.01) x (1 x 0.001) x (6 x 0.0001).  The answer is 3.1416, which is an approximation of Pi (π).

This skill deals with rounding numbers to the nearest 10 or 100 or 1000, etc.  It also helps us understand the concept of place values.

It is important to know, at a glance, the place value of a specific digit.  When working with whole numbers (integers), these place values are assigned by going right-to-left.

The ONES place is the last digit on the right.  Then, moving left, we find the TENS place, then the HUNDREDS place.  The THOUSANDS place is the 4th digit from the right.  Then we find the TEN THOUSANDS place, the HUNDRED THOUSANDS place, and the MILLIONS place. And so on…

When we are working with decimals, all of the previously-named place values are to the LEFT of the decimal.  The decimal place values are assigned by going left-to-right.

The TENTHS place is the first digit after the decimal.  Then, moving to the right, we have the HUNDREDTHS place, then the THOUSANDTHS place. The TEN-THOUSANDTHS place is the 4th digit.  The HUNDRED-THOUSANDTHS place is the 5th digit.  The MILLIONTHS place is the 6th digit, and so on.

When we round to a place value, we check the digit to the right of it.  If it is a 5 or greater, we round up.  If it is 4 or less, we round down.

For example, let us round 864,027 to the nearest TEN THOUSANDS place.  We look at the digit to the right: 4.  Since this digit is 4 or less, we round down.  The answer is 860,000.

Now let us try a decimal.  Round 3.14159 to the nearest THOUSANDTHS place.  We look at the digit to its right: 5.  Since this digit is 5 or greater, we round up.  The answer is 3.142.