2.3: Subtracting Decimals
Introduction
A Question of Times
After winning the Regional Championship, Ms. Sutter took the girls out for ice cream sundaes. While they were sitting at the table, the four girls began talking about all of their times and about how much fun they had running.
“It was great, but I was nervous,” Uniqua said. “especially since I was the last runner.”
“Yes, I can see why you would be, but you did great,” Jessica complimented her.
“It was a close race,” Tasha said.
“Not that close,” Jessica said.
“Sure it was,” Karin chimed in. “It was close between all three winning teams and it was a close call for us to beat last year’s time too.”
“Are you sure?” Jessica questioned.
“Sure I am, look at these times,” Karin said, writing on a napkin.
53.87 Last year’s team
53.73 Our time
53.80 \begin{align*}2^{nd}\end{align*}
53.91 \begin{align*}3^{rd}\end{align*}
“You can figure this out girls, by doing the math,” Ms. Sutter said smiling.
Ms. Sutter is correct. Subtracting decimals is a way to figure out the difference between winning and coming in second or third. What is the difference between last year’s time and this year’s? What is the difference between second and first? What is the difference between first and third?
The best way to figure out these problems is to subtract. This lesson is all about subtracting decimals. By the time you are finished, you will be able to figure out the differences as well.
What You Will Learn
In this lesson you will learn the following skills.
- Subtract decimals with and without rounding
- Estimate decimal differences using front-end estimation
- Identify and apply the inverse property of addition in decimal operations, using numerical and variable expressions.
- Model and solve real-world problems using simple equations involving decimal addition and subtraction.
Teaching Time
I. Subtract Decimals With and Without Rounding
To compare two values, or find the difference in two values, or find out “how many more” we subtract. Anytime we want to find a “left over” value or a “less than” value we subtract. Anytime that you see these key words you can be sure that subtraction is necessary.
You know how to add decimals and estimate sums. Now it is time to learn how to subtract decimals and estimate differences.
How do we subtract decimals?
We subtract decimals the same way we subtract whole numbers—with special care for place value. For instance, when you subtract 571 from 2,462 you make sure to line up the different place values, so that the thousands are subtracted from thousands, hundreds are subtracted from hundreds, and so on.
Example
\begin{align*}& \ 2,462\\
& \underline{- \ \ 571}\\
& \quad 1891\end{align*}
When you subtract two decimals, we do the same thing. We line up the numbers according to place value and use the decimal points as a starting point. If there are missing digits, we can add in zeros to help us keep our work straight. Let’s look at an example.
Example
\begin{align*}& \ \ 18.98\\
& \underline{- \ 4.50}\\
& \ \ 14.48\end{align*}
Notice that the decimal points are lined up and then each digit is matched according to place value. You can see that the zero was added to help us keep each digit in the correct location.
When is it useful to round?
Rounding is useful when estimating or when you need an approximate answer not an exact one.
In what situations might you want to use rounding?
A situation in which we may want to use rounding is if the numbers are too long for pencil calculations, and an approximate answer is sufficient. Also, if the problem itself tells us that we don’t need an exact answer, then we would be able to save time by rounding. The question might use words like "close to" or "approximate".
In these cases, we round the decimals before we subtract. Look for clues in the problem to tell you whether or not to round before subtracting. If you are asked to round, make sure you round to the right place!
Example
Round the numbers to the nearest tenth, then find the difference, 72.953 - 52.418
This problem asks us to round each number to the tenths place before subtracting.
Underline the number we’re rounding to and bold or circle the number directly to the right of it.
We are rounding to the tenths place, so round to the place directly to the right of the decimal place.
The bolded number in the hundredths place is the one to look at when deciding to round up or down.
72.953 \begin{align*}\rightarrow\end{align*}
52.418 \begin{align*}\rightarrow\end{align*}
Now that the numbers are rounded, we line up the decimal places, and subtract.
\begin{align*}& \quad 73.0\\
& \underline{- \; 52.4}\\
& \quad 20.6\end{align*}
While rounding doesn’t give us an exact answer, you can see that it simplifies the subtraction!
2F. Lesson Exercises
- \begin{align*}5.674 - 2.5 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
- Take 5.67 from 12.378
- Round to the nearest tenth and then subtract, 8.356 - 1.258
Take a few minutes to check your work with a partner. Is your work accurate? Correct any mistakes before moving on to the next section.
II. Estimate Decimal Differences Using Front-End Estimation
You probably use estimation—the process of finding an approximate solution to a problem—without even thinking about it. When you figure out how much money you might need to bring to the movies, you use estimation. When you figure out how long you might wait for the bus, or how far it is to the corner store, you’re estimating values.
There are many different techniques for estimating, some of which provide “better” estimates than others. Rounding numbers before performing an operation is one way to find an estimate.
Front-end estimation is another way to estimate.
Good question! Let’s review the steps to using front-end estimation.
The steps to front-end estimation when estimating decimal differences are a little different.
Because subtraction often requires regrouping, we need to subtract the two leftmost digits at the same time. All you do is subtract the two front end—or leftmost—digits.
Here are the steps:
Front-End Estimation for Subtraction:
--Subtract the two (leftmost) digits at the same time
Make a few notes on front-end estimation then continue with the example.
Example
Use front-end estimation to find the difference. 0.55373 - 0.27449
We’re going to use front-end estimation to subtract 0.27449 from 0.55373. Let’s line up our decimal points and take a closer look at the numbers. Then we’ll perform front-end estimation.
0.55373
0.27449
Front-end estimation tells us to subtract the two "front" or leftmost digits at the same time, so we want to subtract the tenths and hundredths places.
\begin{align*}& \quad 0.55\\ &\underline{- \; 0.27}\\ & \quad 0.28\end{align*}
Example
Use front-end estimation to find the difference, 4.032 - 0.82
Let’s begin by lining up the decimal places—adding zeros as necessary—take a close look at the numbers. Then we’ll perform front-end estimation.
4.038
0.820
Front-end estimation tells us to subtract the two front or leftmost digits at the same time, so we want to subtract the ones and the tenths places.
\begin{align*}& \quad 4.0\\ & \underline{- \; 0.8}\\ & \quad 3.2\end{align*}
One of the numbers extended to the thousandths place but because we are using front end estimation, the only numbers we care about for our answer are the two front end numbers.
2G. Lesson Exercises
Use front-end estimation to find the following differences.
- \begin{align*}4.6712 - 1.34 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
- \begin{align*}6.789 - 3.45 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
- \begin{align*}5.023 - .346 = \underline{\;\;\;\;\;\;\;\;}\end{align*}
Take a few minutes to check your work with a partner.
III. Identify and Apply the Inverse Property of Addition in Decimal Operations, Using Numerical and Variable Expressions
The word “inverse” is one that is used frequently in mathematics. In fact, as you get into higher levels of math, you will see the word “inverse” used more often.
What do we mean when we use the word inverse?
The word inverse means opposite. Sometimes numbers are opposites of each other, and sometimes operations are opposites of each other too. Subtraction is called the inverse, or opposite, of addition. If you’ve ever used addition to check subtraction calculations, you already have an idea of the relationship between addition and subtraction.
The inverse property of addition states that for every number \begin{align*}x, x + (-x) = 0\end{align*}. In other words, when you add \begin{align*}x\end{align*} to its opposite, \begin{align*}-x\end{align*} (also known as the additive inverse), the result is zero.
Adding a negative number is the same as subtracting that number.
Let’s put whole numbers in place of \begin{align*}x\end{align*} to make the property clear: \begin{align*}3 + (-3) = 0\end{align*}.
The inverse property of addition may seem obvious, but it has important implications for solving variable equations which aren’t easily solved using mental math—such as variable equations involving decimals.
Remember back to our work with equations and expressions in the last section? Think about an equation. An equation is a number sentence where one expression is equal to another expression. A powerful fact about equations is that if both sides start out equal, and you do the same thing to both sides, they will still be equal. The inverse property of addition says that a number added to its opposite integer will always equal zero. This property can be used to solve equations by eliminating numbers that are 'cluttering up' a variable. By eliminating all of the other numbers near the variable, the equation tells us what value of the variable should make the equation true. It is important to remember, however, that when using the inverse property on one side of an equation to cancel out an unwanted number, we must do the exact same operation on the other side to keep the equation balanced. The example below illustrates this:
Example
\begin{align*}x + 5 & = 15\\ x + 5 - 5 & = 15 - 5 \rightarrow \text{inverse operation suggests we subtract} \ 5 \ \text{from both sides}\\ x & = 10\end{align*}
Notice how, on the left side of the equation, \begin{align*}+ 5 - 5 = 0\end{align*}, leaving the \begin{align*}x\end{align*} alone.
Our answer is that \begin{align*}x\end{align*} is equal to 10.
Now let’s look at an example where the two expressions in the equation are decimals.
Example
\begin{align*}x + 39.517 = 50.281\end{align*}
First, we need to isolate the variable. To do this, we can use the inverse operation. This is an addition problem, so we use subtraction to get the variable by itself. We subtract 39.517 from both sides of the equation.
\begin{align*}x + 39.517 & = 50.281\\ x + 39.517 - 39.517 & = 50.281 - 39.517 \rightarrow \text{inverse operations: subtract} \ 39.517 \ \text{from both sides}\\ x & = 10.764\end{align*}
Notice how we subtracted 39.517 from both sides of the equation. Remember, both expressions on either side of an equation must be equal at all times. Whatever operation you perform to one side of the equation, you must perform to the other side. In this instance, on the left side of the equation, \begin{align*}+ 39.517 - 39.517 = 0\end{align*}, leaving the \begin{align*}x\end{align*} alone on that side. On the right side of the equation, 50.281 - 39.517 gives us the value of \begin{align*}x\end{align*}.
Our answer is that \begin{align*}x\end{align*} is equal to 10.764.
Example
\begin{align*}x - 43.27 = 182.205\end{align*}
First, we need to isolate the variable. To do this, we can use the inverse operation. Notice that this is a subtraction problem. To isolate the variable we are going to use the inverse of subtraction which is addition.
\begin{align*}x - 43.27 & = 182.205\\ x - 43.27 + 43.27 & = 182.205 + 43.27 \rightarrow \text{inverse operations: add} \ 43.27 \ \text{to both sides}\\ x & = 225.475\end{align*}
Notice how we added 43.27 to both sides of the equation. Remember, both expressions on either side of an equation must be equal at all times. Whatever operation you perform to one side of the equation, you must perform to the other side. In this instance on the left side of the equation \begin{align*}-43.27 + 43.27 = 0\end{align*}, leaving the \begin{align*}x\end{align*} alone. On the right side of the equation, 182.205 + 43.27 gives us the value of \begin{align*}x\end{align*}.
Our answer is that \begin{align*}x\end{align*} is equal to 225.475.
2H. Lesson Exercises
Use the inverse property of addition to solve each equation.
- \begin{align*}x+5.678=12.765\end{align*}
- \begin{align*}x-4.32=19.87\end{align*}
- \begin{align*}x+123.578=469.333\end{align*}
Take a few minutes to check your work with a partner.
IV. Model and Solve Real-World Problems Using Simple Equations Involving Decimal Addition and Subtraction
Now we can take what we have learned about decimal addition and subtraction and apply it to some real world problems. After this section, we will return to the girls on the track team and help them to find differences in their times.
Let’s start by looking at how we can use equations to solve problems that involve decimals. Equations help define relationships between different elements in a problem. To successfully write equations, you need to begin by reading each problem carefully, defining the unknown, and assigning the unknown to a variable such as \begin{align*}x\end{align*}. Then you write an equation which describes the mathematical relationship between that variable and the other information in the problem.
An advantage in writing equations is the ability to translate the phrases in the problem into algebraic expressions. It helps to recognize “key words” in problems that indicate which operation to perform. Look at some of the following phrases translated into algebraic expressions.
- the sum of a number and \begin{align*}3 \ \ \ x + 3\end{align*}
- 3 greater than a number \begin{align*}x + 3\end{align*}
- a number increased by \begin{align*}3 \ \ \ x + 3\end{align*}
- 3 less than a number \begin{align*}x - 3\end{align*}
- a number decreased by \begin{align*}3 \ \ \ x - 3\end{align*}
As you practice writing and solving equations, you will become more and more familiar with algebraic phrases and their corresponding mathematical expressions. Once you have an equation, you can use inverse operations to solve for the value of the variable.
Example
Milo’s Pizza uses a total of 283.349 grams of mozzarella every day. Milo uses 96.82 grams more mozzarella than parmesan. How many grams of parmesan does Milo use every day?
Anytime you see the key words “more than,” you know you’re going to write an addition equation. In this problem, we are trying to find out the number of grams of parmesan Milo uses every day. We’ll give that value the variable \begin{align*}x\end{align*}. We know how many grams of mozzarella he uses, and we know the addition relationship between the parmesan and the mozzarella, so we can write the equation \begin{align*}x + 96.82 = 283.349\end{align*}. Now we can solve for the value of \begin{align*}x\end{align*} using inverse operations.
\begin{align*}x + 96.82 & = 283.349\\ x + 96.82 - 96.82 & = 283.349 - 96.82 \rightarrow \text{inverse operation, subtract} \ 96.82 \ \text{from both sides}\\ x & = 186.529\end{align*}
Remember to put the units of measurement in your answer!
The answer is 186.529 grams.
Example
At Saturday’s track meet, Liz ran 1.96 kilometers less than Sonya. Liz ran 1.258 kilometers. How many kilometers did Sonya run?
Anytime you see the key words “less than,” you know you’re going to write a subtraction equation. In this problem, we are trying to find out the number of kilometers Sonya ran. We’ll give that value the variable \begin{align*}x\end{align*}. We know how many kilometers Liz ran, and we know the subtraction relationship between the Liz’s distance and Sonya’s distance, so we can write the equation \begin{align*}x - 1.96 = 1.258\end{align*}. Now we can solve for the value of \begin{align*}x\end{align*} using inverse operations.
\begin{align*}x - 1.96 & = 1.258\\ x - 1.96 + 1.96 & = 1.258 + 1.96 \rightarrow \text{inverse operation, add} \ 1.96 \ \text{to both sides}\\ x & = 3.218\end{align*}
Remember to put the units of measurement in your answer!
The answer is 3.218 kilometers.
Now let’s go back to the original problem and help the girls figure out the differences in track times.
Real Life Example Completed
A Question of Times
Here is the original problem once again. Reread the problem and underline any important information.
After winning the Regional Championship, Ms. Sutter took the girls out for an ice cream sundae. While they were sitting at the table, the four girls began rethinking all of their times and talking about how much fun they had running.
“It was great, but I was nervous,” Uniqua said. “especially since I was the last runner.”
“Yes, I can see why you would be, but you did great,” Jessica complimented her.
“It was a close race,” Tasha said.
“Not that close,” Jessica said.
“Sure it was,” Karin chimed in. “It was close between all three winning teams and it was a close call for us to beat last year’s time too.”
“Are you sure?” Jessica questioned.
“Sure I am, look at these times,” Karin said writing on a napkin.
53.87 Last year’s team
53.73 Our time
53.80 \begin{align*}2^{nd}\end{align*} place
53.91 \begin{align*}3^{rd}\end{align*} place
“You can figure this out girls, by doing the math,” Ms. Sutter said smiling.
Ms. Sutter is correct. Subtracting decimals is a way to figure out the difference between winning and coming in second or third. What is the difference between last year’s time and this year’s? What is the difference between second and first? What is the difference between first and third?
First, we find the difference between last year’s time and this year’s time. We subtract one time from the other. We line up our decimal points and subtract.
\begin{align*}& \quad 53.87\\ & \underline{- \; 53.73}\\ & \qquad .14\end{align*}
The girls beat the previous year’s time by fourteen hundredths of a second. Wow! That is close!
Next, we look at the times between first and second place.
\begin{align*}& \quad 53.80\\ & \underline{- \; 53.73}\\ & \qquad .07\end{align*}
The difference between first and second place was very close! The girls only won by three hundredths of a second.
Finally, we can look at the difference between the first and third place finishes.
\begin{align*}53.91 -53.73 = .18\end{align*}
Wow! The first three teams had very close finishes!!!
Vocabulary
Here are the vocabulary words that are found in this lesson.
- Difference
- a key word that means subtraction.
- Estimate
- to find an approximate answer to a problem.
- Rounding
- one method of estimating where a number is changed according to the place value that it is closest to.
- Front-end Estimation
- a method of estimating. With subtraction, you only subtract the first two places at the same time.
- Inverse
- means opposite
- Inverse Property of Addition
- means that when you add two opposite values together that the answer is 0.
- Inverse Operations
- using the opposite operation can help to solve for an unknown variable in an equation.
Technology Integration
Khan Academy Subtracting Decimals
James Sousa, Subtraction of Decimals
James Sousa, Solving One Step Equations by Adding and Subtracting Whole Numbers
Other Videos:
http://www.schooltube.com/video/05733ec4ba5143739f6d/Adding-and-Subtracting-Decimals – This is a great video on adding and subtracting decimals.
Time to Practice
Directions: Find each difference.
1. 6.57 - 5.75
2. .0826 - .044
3. 19.315 - 6.8116
4. 2056.04 - 2044.1
5. 303.45 - 112.05
6. 16.576 - 8.43
7. 199.2 - 123.45
8. 1.0009 - .234
9. 789.12 - .876
10. 102.03 - .27
Directions: Find the difference after rounding each decimal to the nearest hundredth.
11. 63.385 - 50.508
12. .361 - .295
13. 747.005 - 47.035
14. .882 - .596
15. .9887 - .0245
Directions: Use front-end estimation to estimate the following differences.
16. .22281 - .1946
17. 4.033 - .8821
18. .6252 - .4406
19. 3.522 - 1.093
20. 5.678 - 2.93
Directions: Solve each equation for the missing variable using the inverse operation.
21. Solve \begin{align*}x + 4.39 = 7.22\end{align*}
22. Solve \begin{align*}818.703 = 614.208 + x\end{align*}
23. Solve \begin{align*}x + 55.27 = 100.95\end{align*}
Directions: Use equations to solve each word problem. Each answer should have an equation and a value for the variable.
24. Jamal’s leek and potato soup calls for 2.45 kg more potatoes than leeks. Jamal uses 4.05 kg of potatoes. How many kilograms of leeks does he use? Write an equation and solve.
25. He distance from Waterville to Longford is 118.816 kilometers less than the distance from Transtown to Longford. The distance from Waterville to Longford is 67.729 kilometers. What is the distance from Transtown to Longford? Write an equation and solve.
26. Sabrina spent $25.62 at the book fair. When she left the fair, she had $6.87. How much did money did she take to the fair? Write an equation and solve.
27. Mr. Bodin has 11.09 liters of a cleaning solution, which is a combination of soap and water. If there are 2.75 liters of soap in the solution, how many liters of water are in the solution? Write an equation and solve.
Notes/Highlights Having trouble? Report an issue.
Color | Highlighted Text | Notes | |
---|---|---|---|
Please Sign In to create your own Highlights / Notes | |||
Show More |