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# 1.2: Subtraction of Real Numbers

Difficulty Level: At Grade Created by: CK-12

## Subtraction of Integers

Objectives

The lesson objectives for The Subtraction of Real Numbers are:

• Subtraction of Integers Using Models
• Subtraction of Integers Using the Number Line
• Subtraction of Integers Using the Rules

Introduction

In this concept you will learn to subtract integers by using different representations. You will learn how to subtract integers by using appropriate models and by using the number line. These methods will lead to the formation of two rules for subtracting integers.

Watch This

Guidance

Example A

• To subtract one signed number from another, change the question from a subtraction question to an addition question, and change the sign of the number that was originally being subtracted. In other words, to subtract signed numbers simply add the opposite. Once these changes have been made, follow the rules for adding signed numbers.

\begin{align*}7-(-3)=? \quad 7+(+3)=?\end{align*}

The problem can be represented by using color counters. In this case, the red counters represent positive numbers.

The above representation shows the addition of 7 positive counters and 3 positive counters.

The answer is the sum of 7 and 3. The answer takes on the sign of the two digits being added and in this case the digits both have a positive value. Therefore, \begin{align*}7+(+3)=10\end{align*}

Example B

\begin{align*}4-(+6) &= ?\\ 4+(-6) &= ?\end{align*}

Change the problem to an addition problem and change the sign of the original number that was being subtracted.

The above representation shows the addition of 4 positive counters and 6 negative counters.

One positive counter and one negative counter equals zero. \begin{align*}1+(-1)=0\end{align*}

Draw a line through the counters that equal zero.

The remaining counters represent the answer. Therefore, \begin{align*}4-(+6)=-2\end{align*}. The answer is the difference between 6 and 4. The answer takes on the sign of the larger digit and in this case the six has a negative value and it is greater than 4.

Example C

This same method can be extended to adding variables. Algebra tiles can be used to represent positive and negative values.

\begin{align*}5x-(+8x) &= ?\\ 5x+(-8x) &= ?\end{align*}

The green algebra tiles represent positive \begin{align*}x\end{align*} and the white tiles represent negative \begin{align*}x\end{align*}. There are 5 positive \begin{align*}x \ tiles\end{align*} and 8 negative \begin{align*}x \ tiles\end{align*}.

The remaining algebra tiles represent the answer. There are three negative \begin{align*}x\end{align*} tiles remaining. Therefore, \begin{align*}(6x)-(+8x)=-3x\end{align*}. The answer is the difference between \begin{align*}8x\end{align*} and \begin{align*}5x\end{align*}. The answer takes on the sign of the larger digit and in this case the eight has a negative value and it is greater than 5.

Example D

\begin{align*}(-4)-(+3) &= ?\\ (-4)+(-3) &= ?\end{align*}

The solution to this problem can be determined by using the number line.

Indicate the starting point of -4 by using a dot. From this point, add a -3 by moving three places to the left. You will stop at -7.

The point where you stopped is the answer to the problem. Therefore, \begin{align*}(-4)-(+3)=-7\end{align*}

The answer is the sum of 4 and 3. The answer takes on the sign of the digits being added. In this case the 4 and the 3 have negative signs. The answer will be a negative number.

From using models to subtract integers, there are two rules that become obvious. These rules are:

• When you subtract integers you change the question to an addition problem and change the sign of the original number being subtracted.
• Follow the rules for adding integers. When you add two integers with the same sign, add the numbers and use the sign of the digits being added. When you add two integers that have opposite signs, subtract the numbers and use the sign of the larger digit.

Vocabulary

Integer
All natural numbers, their opposites, and zero are integers. A number in the list ..., -3, -2, -1, 0, 1, 2, 3...
Number Line
A number line is a line that matches a set of points and a set of numbers one to one.

It is often used in mathematics to show mathematical computations.

Guided Practice

1. Use a model to answer the problem \begin{align*}(-2)-(-6)=?\end{align*}
2. Use the number line to determine the answer to the problem \begin{align*}7-(+5)=?\end{align*}
3. Determine the answer to \begin{align*}(-8)-(-5)=?\end{align*} and \begin{align*}(-4)-(+9)=?\end{align*} by using the rules for subtracting integers.

1. \begin{align*}(-2)-(-6) &= ?\\ (-2)+(+6) &= ?\end{align*}

Cancel the counters that equal zero

There are 4 positive counters left. Therefore, \begin{align*}(-2)-(-6)=4\end{align*}. The answer is the difference between 6 and 2. The answer takes on the sign of the larger digit and in this case the six has a positive value and it is greater than 2.

2. \begin{align*}(7)-(+5) &=?\\ (7)+(-5) &= ?\end{align*}

You begin on 7 and move five places to the left. You stop at 2. Therefore \begin{align*}7-(+5)=2\end{align*}.

The answer is the difference between 7 and 5. The answer takes on the sign of the larger digit and in this case the seven has a positive value and it is greater than 5.

3. \begin{align*}(-8)-(-5) &= ?\\ (-8)+(+5) &= ?\end{align*}

When the problem is written as an addition question and the sign of the original number being subtracted is changed to a positive, the numbers have opposite signs. The numbers must be subtracted and the answer is a negative value since the larger digit of 8 has a negative sign.. Therefore, \begin{align*}(-8)-(-5)=-3\end{align*}

\begin{align*}(-4)-(+9) &= ?\\ (-4)+(-9) &= ?\end{align*}

When the problem is written as an addition question and the sign of the original number being subtracted is changed to a negative, the numbers have the same signs. The numbers must be added and the answer is a negative value. Therefore, \begin{align*}(-4)-(+9)=-13\end{align*}

Summary

The subtraction of integers can be represented with manipulatives such as color counters and algebra tiles. A number line can also be used to show the subtraction of integers. The subtraction of integers can be done by following two rules: When you subtract integers you change the question to an addition problem and change the sign of the original number being subtracted. When the problem has been written as an addition problem, follow the rules for adding integers. When you add two integers with the same sign, add the numbers and use the sign of the digits being added. When you add two integers that have opposite signs, subtract the numbers and use the sign of the larger digit.

Problem Set

Use color counters to represent the following subtraction problems and use that model to determine the answer.

1. \begin{align*}(-9)-(-2)\end{align*}
2. \begin{align*}(5)-(+8)\end{align*}
3. \begin{align*}(5)-(-4)\end{align*}
4. \begin{align*}(-7)-(-9)\end{align*}
5. \begin{align*}(6)-(+5)\end{align*}

Use a number line to represent the following subtraction problems and use the number line to determine the answer.

1. \begin{align*}(8)-(+4)\end{align*}
2. \begin{align*}(-2)-(-7)\end{align*}
3. \begin{align*}(3)-(+5)\end{align*}
4. \begin{align*}(-6)-(-10)\end{align*}
5. \begin{align*}(-4)-(-7)\end{align*}

Use the rules that you have learned for subtracting integers to answer the following problems and state the rule that you used.

1. \begin{align*}(-13)-(-19)\end{align*}
2. \begin{align*}(-6)-(+8)-(-12)\end{align*}
3. \begin{align*}(14)-(+8)-(-6)\end{align*}
4. \begin{align*}(18)-(+8)-(+3)\end{align*}
5. \begin{align*}(10)-(-6)-(+4)-(+2)\end{align*}

For each of the following models, write a subtraction problem and answer the problem. (Hint: You may find it easier to write an addition problem and then to rewrite the problem as a subtraction question)

Use color counters...

\begin{align*}& (-9)-(-2)\\ & (-9)+(+2)\end{align*}

Two negative counters cancel the two positive counters.

There are 7 negative counters remaining. Therefore \begin{align*}(-9)-(-2)=-7\end{align*}.

\begin{align*}& (5)-(-4)\\ & (5)+(+4)\end{align*}

There are no negative counters to cancel the positive counters. There are 9 positive counters.

Therefore \begin{align*}(5)-(-4)=9\end{align*}

\begin{align*}& (6)-(+5)\\ & (6)+(-5)\end{align*}

Five positive counters cancel out 5 negative counters. There is 1 positive counter remaining.

Therefore \begin{align*}(6)-(+5)=1\end{align*}

Use a number line...

1. \begin{align*}(8)-(+4)\end{align*} \begin{align*}(8)+(-4)=4\end{align*}
1. \begin{align*}(3)-(+5)\end{align*} \begin{align*}(3)+(-5)=-2\end{align*}
1. \begin{align*}(-4)-(-7)\end{align*} \begin{align*}(-4)+(+7)=3\end{align*}

Use the rules...

\begin{align*}& (-13)-(-9)\\ & (-13)+(+9)=-4\end{align*}

The problem was first written as an addition problem and the sign of the number being subtracted was changed. The numbers have unlike or opposite signs so they must be subtracted. The answer has a negative sign which is the same sign as the larger digit 13.

\begin{align*}& (14)-(+8)-(-6)\\ & (14)+(-8)+(+6)=(20)+(-8)=12\end{align*}

The problem was first written as an addition problem and the sign of the number being subtracted was changed. The two positive numbers were added and the answer of 20 has a positive sign which is the sign of the numbers being added. The answer of 20 was then added to -8 and the answer was calculated by subtracting the two numbers. The answer of 12 has a positive sign which is the sign of the larger number 20.

\begin{align*}& (10)-(-6)-(+4)-(+2)\\ & (10)+(+6)+(-4)+(-2)\\ & 10+6=16\\ & -4+(-2)=-6\\ & 16+(-6)=10\end{align*}

The problem was first written as an addition problem and the sign of the numbers being subtracted were changed. The two positive numbers were added to give a positive answer of 16. The two negative numbers were added to give a negative answer of -6. The two answers were then subtracted to give the final answer of 10. The answer was calculated by subtracting the two numbers and applying the positive sign of the larger number to the answer.

For each of the following models...

1. \begin{align*}& (-9x)-(-4x)\\ & (-9x)+(+4x)\\ & =-5x\end{align*}
1. \begin{align*}& (4x^2)-(+2x^2)-(-2x)-(+4x)\\ & (4x^2)+(-2x^2)+(+2x)+(-4x)\\ & =2x^2-2x\end{align*}
1. \begin{align*}& (7x)-(2x)\\ & (7x)+(-2x)\\ & =5x\end{align*}

## Subtraction of Fractions

Objectives

The lesson objectives for The Subtraction of Real Numbers are:

• Subtraction of Fractions Using Models
• Subtraction of Fractions Using the Number Line
• Subtraction of Fractions Using the Rules

Introduction

In this concept you will learn to subtract real numbers using different representations. You will learn to subtract fractions by using appropriate models and by using the number line. These methods will lead to the formation of rules for subtracting fractions.

Watch This

Guidance

\begin{align*}\frac{5}{7}-\frac{2}{7}=?\end{align*}

The problem can be represented by using fraction strips. You can create these fraction strips yourself or you can use commercial pieces called Fraction Factory pieces. Those being presented in the following examples are not the commercial type. Therefore, the colors used are simply a personal choice. This first example will explore subtracting positive fractions that have the same denominator.

\begin{align*}\boxed{\frac{5}{7}-\frac{2}{7}=\frac{5-2}{7}=\frac{3}{7}}\end{align*}

To subtract fractions, the fractions must have the same bottom numbers (denominators). Both fractions have a denominator of 7. The answer is the result of subtracting the top numbers (numerators). The numbers in the numerator are 5 and 2. The difference of 5 and 2 is 3. This difference is written in the numerator over the denominator of 7. Therefore \begin{align*}\frac{5}{7}-\frac{2}{7}=\frac{3}{7}\end{align*}.

Example A

\begin{align*}\frac{8}{11}-\frac{6}{11}=?\end{align*}

This first example will explore subtracting positive fractions that have the same denominator.

\begin{align*}\boxed{\frac{8}{11}-\frac{6}{11}=\frac{8-6}{11}=\frac{2}{11}}\end{align*}

To subtract fractions, the fractions must have the same bottom numbers (denominators). Both fractions have a denominator of 11. The answer is the result of subtracting the top numbers (numerators). The numbers in the numerator are 8 and 6. The difference of 8 and 6 is 2. This difference is written in the numerator over the denominator of 11. Therefore \begin{align*}\frac{8}{11}-\frac{6}{11}=\frac{2}{11}\end{align*}.

Example B

Bessie is measuring the amount of soda in the two coolers in the cafeteria. She estimates that the first cooler is \begin{align*}\frac{2}{3}\end{align*} full and the second cooler is \begin{align*}\frac{1}{4}\end{align*} full. What single fraction could Bessie use to represent how much more soda is in the first cooler than in the second cooler?

Use fraction strips to represent each fraction.

\begin{align*}\frac{2}{3}\end{align*} and \begin{align*}\frac{8}{12}\end{align*} are equivalent fractions. \begin{align*}\frac{2}{3} \left(\frac{4}{4}\right)=\frac{8}{12}\end{align*}.

\begin{align*}\frac{1}{4}\end{align*} and \begin{align*}\frac{3}{12}\end{align*} are equivalent fractions. \begin{align*}\frac{1}{4} \left(\frac{3}{3}\right)=\frac{3}{12}\end{align*}.

The two green pieces will be replaced with eight purple pieces and the one blue piece will be replaced with three purple pieces.

The denominator of 12 is the LCD (least common denominator) of \begin{align*}\frac{2}{3}\end{align*} and \begin{align*}\frac{1}{4}\end{align*} because it is the LCM (least common multiple) of the denominators 3 and 4.

Example C

A number line can also be used to subtract fractions. In the following example, a mixed number which is a whole number and a fraction will be added to a fraction by using a \begin{align*}\frac{1}{4}\end{align*} number line.

\begin{align*}1 \frac{3}{4}-\frac{1}{2}\end{align*}

The number line is labeled in intervals of 4 which indicates that each interval represents \begin{align*}\frac{1}{4}\end{align*}. From zero, move to the number 1 plus 3 more intervals to the right. Mark the location. This represents \begin{align*}1 \frac{3}{4}\end{align*}.

From here, move to the left \begin{align*}\frac{1}{2}\end{align*} or \begin{align*}\frac{1}{2}\end{align*} of 4, which is 2 intervals. An equivalent fraction for \begin{align*}\frac{1}{2}\end{align*} is \begin{align*}\frac{2}{4}\end{align*}.

The difference of \begin{align*}1 \frac{3}{4}\end{align*} and \begin{align*}\frac{1}{2}\end{align*} is \begin{align*}1 \frac{1}{4}\end{align*}.

Having used models to subtract fractions, there are two rules that become obvious. These rules are:

• Fractions can be subtracted if they have the same denominator. To subtract fractions that have the same denominator, subtract the numerators and write the difference over the common denominator.
• When you subtract fractions that have different denominators, you must express the fractions as equivalent fractions with a LCD. Now, subtract the numerators and write the difference over the common denominator.

Vocabulary

Denominator
The denominator of a fraction is the number on the bottom that indicates the total number of equal parts in the whole or the group. \begin{align*}\frac{5}{8}\end{align*} has denominator 8.
Fraction
A fraction is any rational number that is not an integer.
LCD
The least common denominator is the lowest common multiple of the denominators of two or more fractions. The least common denominator of \begin{align*}\frac{3}{4}\end{align*} and \begin{align*}\frac{1}{5}\end{align*} is 20.
LCM
The least common multiple is the lowest common multiple that two or more numbers share. The least common multiple of 6 and 5 is 30.
Numerator
The numerator of a fraction is the number on top that is the number of equal parts being considered in the whole or the group. \begin{align*}\frac{5}{8}\end{align*} has 'numerator 5.

Guided Practice

1. Use a model to answer the problem \begin{align*}\frac{7}{10}-\frac{2}{5}=?\end{align*}
2. Use a number line to determine the answer to the problem \begin{align*}\frac{7}{8}-\frac{1}{2}\end{align*}.
3. Determine the answer to \begin{align*}\frac{5}{8}-\frac{1}{3}=?\end{align*} and \begin{align*}\frac{4}{5}-\frac{1}{4}=?\end{align*} by using the rules for subtracting fractions.

1.

Use fraction strips to represent each fraction.

The fractions do not have a common denominator. An equivalent fraction for \begin{align*}\frac{2}{5}\end{align*} is \begin{align*}\left(\frac{2}{2}\right) \left(\frac{2}{5}\right)=\frac{4}{10}\end{align*}

When four strips have been removed from \begin{align*}\frac{7}{10}\end{align*}, there are \begin{align*}\frac{3}{10}\end{align*} remaining.

\begin{align*}\boxed{\frac{7}{10}-\frac{4}{10}=\frac{7-4}{10}=\frac{3}{10}}\end{align*}

2. \begin{align*}\frac{7}{8}-\frac{1}{2}\end{align*}

Use a \begin{align*}\frac{1}{8}\end{align*} number line. The number line is labeled in intervals of 8. Place the starting point at \begin{align*}\frac{7}{8}\end{align*}.

From this point, move to the left a total of 4 intervals. \begin{align*}\frac{1}{2}\end{align*} of \begin{align*}8=4\end{align*}. An equivalent fraction for \begin{align*}\frac{1}{2}\end{align*} is \begin{align*}\frac{1}{2} \left(\frac{4}{4} \right)=\frac{4}{8}\end{align*}. The point where you stop is the difference of \begin{align*}\frac{7}{8}\end{align*} and \begin{align*}\frac{1}{2}\end{align*}.

On the number line you stopped at the point \begin{align*}\frac{3}{8}\end{align*}. This is equal to \begin{align*}\boxed{\frac{7}{8}-\frac{4}{8}=\frac{7-4}{8}=\frac{3}{8}}\end{align*}.

3. \begin{align*}\frac{5}{8}-\frac{1}{3}=?\end{align*}

The least common multiple of 8 and 3 is 24. This means that both fractions must have a common denominator of 24 before they can be subtracted.

\begin{align*}\frac{5}{8} \left(\frac{3}{3}\right)=\frac{15}{24}\end{align*} \begin{align*}\frac{5}{8}\end{align*} and \begin{align*}\frac{15}{24}\end{align*} are equivalent fractions.

\begin{align*}\frac{1}{3} \left(\frac{8}{8}\right)=\frac{8}{24}\end{align*} \begin{align*}\frac{1}{3}\end{align*} and \begin{align*}\frac{8}{24}\end{align*} are equivalent fractions.

\begin{align*}& \frac{5}{8}-\frac{1}{3}\\ & \frac{15}{24}-\frac{8}{24}\\ & =\frac{15-8}{24}=\frac{7}{24}\end{align*}

\begin{align*}\frac{4}{5}-\frac{1}{4}=?\end{align*}

The least common multiple of 5 and 4 is 20. This means that both fractions must have a common denominator of 20 before they can be added.

\begin{align*}\frac{4}{5} \left(\frac{4}{4}\right)=\frac{16}{20}\end{align*} \begin{align*}\frac{4}{5}\end{align*} and \begin{align*}\frac{16}{20}\end{align*} are equivalent fractions.

\begin{align*}\frac{1}{4} \left(\frac{5}{5}\right)=\frac{5}{20}\end{align*} \begin{align*}\frac{1}{4}\end{align*} and \begin{align*}\frac{5}{20}\end{align*} are equivalent fractions.

\begin{align*}& \frac{4}{5}-\frac{1}{4}\\ & \frac{16}{20}-\frac{5}{20}\\ & =\frac{16-5}{20}=\frac{11}{20}\end{align*}

Summary

The subtraction of fractions can be represented with a manipulative such as a fraction strip. A number line can also be used to show the subtraction of fractions.

The subtraction of fractions can be done by following two rules:

Fractions can be subtracted only if they have the same denominator. To subtract fractions that have the same denominator, subtract the numerators and write the difference over the common denominator.

In order to subtract fractions that have different denominators, the fractions must be expressed as equivalent fractions with a LCD. The difference of the numerators can be written over the common denominator.

Problem Set

Use fraction strips to represent the following subtraction problems and use that model to determine the answer.

1. \begin{align*}\frac{3}{4}-\frac{5}{8}\end{align*}
2. \begin{align*}\frac{4}{5}-\frac{2}{3}\end{align*}
3. \begin{align*}\frac{5}{9}-\frac{2}{3}\end{align*}
4. \begin{align*}\frac{6}{7}-\frac{2}{3}\end{align*}
5. \begin{align*}\frac{7}{10}-\frac{1}{5}\end{align*}

Use a number line to represent the following subtraction problems and use the number line to determine the answer.

1. \begin{align*}\frac{2}{3}-\frac{1}{2}\end{align*}
2. \begin{align*}\frac{3}{5}-\frac{3}{10}\end{align*}
3. \begin{align*}\frac{7}{9}-\frac{1}{3}\end{align*}
4. \begin{align*}\frac{5}{8}-\frac{1}{4}\end{align*}
5. \begin{align*}\frac{2}{5}-\frac{2}{10}\end{align*}

Use the rules that you have learned for subtracting fractions to answer the following problems.

1. \begin{align*}\frac{7}{11}-\frac{1}{2}\end{align*}
2. \begin{align*}\frac{5}{8}-\frac{5}{12}\end{align*}
3. \begin{align*}\frac{5}{6}-\frac{3}{4}\end{align*}
4. \begin{align*}\frac{5}{6}-\frac{2}{5}\end{align*}
5. \begin{align*}\frac{4}{5}-\frac{3}{4}\end{align*}

For each of the following questions, write a subtraction statement and find the result.

1. Sally used \begin{align*}\frac{2}{3} \ cups\end{align*} of flour to make cookies. Terri used \begin{align*}\frac{1}{2} \ cups\end{align*} of flour to make a cake. Who used more flour? How much more flour did she use?
2. Lauren used \begin{align*}\frac{3}{4} \ cup\end{align*} of milk, \begin{align*}1 \frac{1}{3} \ cups\end{align*} of flour and \begin{align*}\frac{3}{8} \ cup\end{align*} of oil to make pancakes. Alyssa used \begin{align*}\frac{3}{8} \ cup\end{align*} of milk, \begin{align*}2 \frac{1}{4} \ cups\end{align*} of flour and \begin{align*}\frac{1}{3} \ cup\end{align*} of melted butter to make waffles. Who used more cups of ingredients? How many more cups of ingredients did she use?
3. Write two fractions with different denominators whose difference is \begin{align*}\frac{3}{8}\end{align*}. Use fraction strips to model your answer.
4. Jake’s dog ate \begin{align*}12 \frac{2}{3} \ cans\end{align*} of food in one week and \begin{align*}9 \frac{1}{4} \ cans\end{align*} the next week. How many more cans of dog food did Jake’s dog eat in week one?
5. Sierra and Clark each solved the same problem. Sierra’s Solution \begin{align*}& \frac{3}{4}-\frac{1}{6}\\ & \frac{9}{12}-\frac{2}{12}\\ & =\frac{7}{12}\end{align*} Clark’s Solution \begin{align*}& \frac{3}{4}-\frac{1}{6}\\ & \frac{9}{12}-\frac{2}{12}\\ & =\frac{7}{0}\end{align*} Who is correct? What would you tell the person who has the wrong answer?

Use fraction strips...

1. \begin{align*}\frac{3}{4}-\frac{5}{8}\end{align*} The strips have been changed so that both have a common denominator of eight. Remove 5 strips from the 6 strips to determine the answer. \begin{align*}\boxed{\frac{3}{4}-\frac{5}{8}=\frac{6}{8}-\frac{5}{8}}\end{align*} \begin{align*}\boxed{\frac{6}{8}-\frac{5}{8}=\frac{6-5}{8}=\frac{1}{8}}\end{align*}
1. \begin{align*}\frac{5}{9}-\frac{1}{3}\end{align*} The strips have been changed so that both have a common denominator of nine. The least common multiple of 9 and 3 is 9. Therefore both fractions must have a common denominator of 9. \begin{align*}\frac{1}{3} \left(\frac{3}{3}\right)=\frac{3}{9}\end{align*}. Three strips must be removed from the 5 strips to determine the answer. \begin{align*}& \frac{5}{9}-\frac{1}{3}\\ & \frac{5}{9}-\frac{3}{9}\\ & \frac{5-3}{9}=\frac{2}{9}\end{align*}
1. \begin{align*}\frac{7}{10}-\frac{1}{5}\end{align*} \begin{align*}\boxed{\frac{7}{10}-\frac{2}{10}}\end{align*} The least common multiple of 10 and 5 is 10. Therefore both fractions must have a common denominator of 10. \begin{align*}\frac{1}{5} \left(\frac{2}{2}\right)=\frac{2}{10}\end{align*}. When 2 strips are removed from the 7 strips, the answer is \begin{align*}\frac{5}{10}\end{align*} which is equivalent to \begin{align*}\frac{1}{2}\end{align*}. \begin{align*}& \frac{7}{10}-\frac{1}{5}\\ & \frac{7}{10}-\frac{2}{10}\\ & \frac{7-2}{10}=\frac{5}{10}\\ & \frac{5}{10}=\frac{1}{2}\end{align*}

Use a number line...

1. \begin{align*}\frac{2}{3}-\frac{1}{2}\end{align*} The least common multiple of 3 and 2 is 6. Use a number line that is marked in intervals of 6. \begin{align*}\frac{2}{3} & \rightarrow \frac{2}{3} \left(\frac{2}{2}\right)=\frac{4}{6}\\ \frac{1}{2} & \rightarrow \frac{1}{2} \left(\frac{3}{3}\right)=\frac{3}{6}\end{align*} Begin by placing a dot above the \begin{align*}4^{th}\end{align*} interval and then move 3 places to the left. You will be at your answer. \begin{align*}& \frac{2}{3}-\frac{1}{2}\\ & \frac{4}{6}-\frac{3}{6}\\ & \frac{4-3}{6}=\frac{1}{6}\end{align*}
1. \begin{align*}\frac{7}{9}-\frac{1}{3}\end{align*} Use a \begin{align*}\frac{1}{9}\end{align*} number line since the least common multiple of 9 and 3 is 9. \begin{align*}\frac{2}{3} \rightarrow \frac{1}{3} \left(\frac{3}{3}\right)=\frac{3}{9}\end{align*} \begin{align*}& \frac{7}{9}-\frac{1}{3}\\ & \frac{7}{9}-\frac{3}{9}\\ & \frac{7-3}{9}=\frac{4}{9}\end{align*}
1. \begin{align*}\frac{2}{5}-\frac{2}{10}\end{align*} The least common multiple of 5 and 10 is 10. Use a number line that is labeled in intervals of 10. \begin{align*}\frac{2}{5} \rightarrow \frac{2}{5} \left(\frac{2}{2}\right)=\frac{4}{10}\end{align*} \begin{align*}& \frac{2}{5}-\frac{2}{10}\\ & \frac{4}{10}-\frac{2}{10}\\ & \frac{2}{10}=\frac{1}{5}\end{align*}

Use the rules...

1. \begin{align*}\frac{7}{11}-\frac{1}{2}\end{align*} The least common multiple of 11 and 2 is 22. \begin{align*}& \frac{7}{11} \left(\frac{2}{2}\right)-\frac{1}{2} \left(\frac{11}{11}\right)\\ & \frac{14}{22}-\frac{11}{22}\\ & =\frac{3}{22}\end{align*}
1. \begin{align*}\frac{5}{6}-\frac{3}{4}\end{align*} The least common multiple of 4 and 6 is 12. \begin{align*}& \frac{5}{6} \left(\frac{2}{2}\right)-\frac{3}{4} \left(\frac{3}{3}\right)\\ & \frac{10}{12}-\frac{9}{12}\\ & =\frac{1}{12}\end{align*}
1. \begin{align*}\frac{4}{5}-\frac{3}{4}\end{align*} The least common multiple of 5 and 4 is 20. \begin{align*}& \frac{4}{5} \left(\frac{4}{4}\right)-\frac{3}{4} \left(\frac{5}{5}\right)\\ & \frac{16}{20}-\frac{15}{20}\\ & =\frac{1}{20}\end{align*}

For each of the following questions...

1. \begin{align*}\frac{2}{3}\end{align*} and \begin{align*}\frac{1}{2}\end{align*} The least common multiple of 3 and 2 is 6. Change both fractions to the same denominator to determine who used the most flour. \begin{align*}\frac{2}{3} \left(\frac{2}{2}\right)&= \frac{4}{6}\\ \frac{1}{2} \left(\frac{3}{3}\right) &= \frac{3}{6}\end{align*} Sally used more flour than Terri since \begin{align*}& \frac{4}{6} > \frac{3}{6}\\ & \frac{4}{6}-\frac{3}{6}\\ & \frac{4-3}{6}\\ & =\frac{1}{6}\end{align*}

Sally used \begin{align*}\frac{1}{6}\end{align*} more cups of flour than Terri.

1. To obtain an answer of \begin{align*}\frac{3}{8}\end{align*} by subtracting two fractions, the fractions with the denominator of 8 could be \begin{align*}\frac{4}{8}\end{align*} and \begin{align*}\frac{1}{8}\end{align*}. Two fractions with different denominators whose difference is \begin{align*}\frac{3}{8}\end{align*} are \begin{align*}\frac{1}{2}\end{align*} and \begin{align*}\frac{1}{8}\end{align*}.
1. Sierra’s Solution \begin{align*}& \frac{3}{4} - \frac{1}{6}\\ & \frac{9}{12}-\frac{2}{12}\\ & =\frac{7}{12}\end{align*} Clark’s Solution \begin{align*}& \frac{3}{4}-\frac{1}{6}\\ & \frac{9}{12}-\frac{2}{12}\\ & =\frac{7}{0}\end{align*} Sierra’s solution is correct. I would tell Clark that when fractions are subtracted, the fractions must have a common denominator. The difference of the numerators is placed over the common denominator. The denominators are NOT subtracted.

## Subtraction of Decimal Numbers

Objectives

The lesson objectives for The Subtraction of Real Numbers are:

• Subtraction of Positive Decimal Numbers.
• Subtraction of Positive and Negative Decimal Numbers
• Subtraction of Decimals Using the Rules

Introduction

In this concept you will learn to subtract decimal numbers. You will learn first to subtract decimal numbers that are positive values by applying the vertical alignment method. Then, you will subtract decimal numbers that are both negative and positive values. You will again apply the rules for adding integers since subtracting decimals that are signed numbers is the same as adding the opposite. Mastering these concepts will lead to the formation of rules for subtracting decimal numbers.

Watch This

Guidance

Jeremy and his family are driving to visit his grandparents. On the first day they drove 234.8 miles and on the second day they drove 251.6 miles. How many more miles did they drive on the second day?

To answer this question, the decimal numbers should be written using the vertical alignment method. The decimal number of greater magnitude should be placed above the number of smaller magnitude. Magnitude is simply the size of the number without respect to its sign or direction. The number -42.8 has a direction to the left and a magnitude of 42.8. The decimal points must be kept directly under each other as well as the digits must be kept in the same place value in line with each other. This means that digits in the ones place must be directly below digits in the ones place, digits in the tenths place must be in the tenths column, digits in the hundredths place must be in the hundredths column and so on. Once the numbers have been correctly aligned, the subtraction process is the same as subtracting whole numbers.

The decimal number 251.6 is of greater magnitude than 234.8. The numbers must be vertically aligned with the larger one above the smaller one. Now the numbers can be subtracted.

They drove 16.8 miles more on the second day.

Example A

Subtract: \begin{align*}57.62 - 6.18\end{align*}

Subtracting decimals is similar to subtracting whole numbers. We line up the decimal points so that we can subtract corresponding place value digits (e.g. tenths from tenths, hundredths from hundredths, and so on). As with whole numbers, we start from the right and work toward the left remembering to borrow when it is necessary.

To answer this question, the decimal numbers should be written using the vertical alignment method. Don’t forget to put the number of greater magnitude above the smaller number.

\begin{align*}& \quad 57. \cancel{\overset{5}{6}} \ ^1 2\\ & \underline{ \; \; -6.1 \; \; \; 8}\\ & \quad 51.4 \ \ 4\end{align*}

Example B

\begin{align*}(98.04)-(32.801)\end{align*}

Begin by writing the question using the vertical alignment method.

\begin{align*}& \quad 98.04\\ & \underline{-32.801}\\ & \end{align*}

The decimal points must be kept directly under each other as well as the digits must be kept in the same place value in line with each other. This means that digits in the ones place must be directly below digits in the ones place, digits in the tenths place must be in the tenths column, digits in the hundredths place must be in the hundredths column and so on. To ensure that the digits are aligned correctly, add zero to 98.04.

\begin{align*}& \ \ 98.04 {\color{blue}0}\\ & \underline{-32.801}\\ & \end{align*}

Subtract the numbers.

\begin{align*}& \ \ 9 \overset{7}{\cancel{8}}.^1 0 \overset{3}{\cancel{4}} \ ^1 {\color{blue}0}\\ & \underline{-32. \; 80 \; \; \; 1}\\ & \ \ 65. \ 23 \quad 9\end{align*}

Example C

\begin{align*}(67.65)-(-25.43)\end{align*}

The first step is to write the problem as an addition problem and to change the sign of the original number being subtracted. In other words, add the opposite.

\begin{align*}(67.65)+(+25.43)\end{align*}

Now, write the problem using the vertical alignment method. The two decimal numbers that are being added have positive signs. Apply the same rule that you used when adding integers that had same signs – add the numbers and use the sign of numbers in the answer.

\begin{align*}& \ \quad 67.65\\ & \underline{ \;\; +25.43}\\ & +93.08\end{align*}

Example D

\begin{align*}(137.4)-(+259.687)\end{align*}

The first step is to write the problem as an addition problem and to change the sign of the original number being subtracted. In other words, add the opposite.

\begin{align*}(137.4)+(-259.687)\end{align*}

Now write the problem using the vertical alignment method. Remember to put 259.687 above 137.4 because 259.687 is the number of greater magnitude. The two decimal numbers that are being added have opposite signs. Apply the same rule that you used when adding integers that had opposite signs – subtract the numbers and use the sign of the larger number in the answer.

\begin{align*}&-259.687\\ & \underline{ \; +137.4 \;\;\;}\\ & \end{align*}

To ensure that the digits are aligned correctly, add zeros to 137.4.

\begin{align*}& -259.687\\ & \underline{ \; +137.4 {\color{blue}00}}\\ & \end{align*}

Subtract the numbers.

\begin{align*}& \ -259.687\\ & \underline{ \;\; +137.4 {\color{blue}00}}\\ & -122.287\end{align*}

The decimal numbers being added have opposite signs. This means that the sign of the answer will be the same sign as that of the number of greater magnitude. In this problem the answer will have a negative sign.

Vocabulary

Decimal Number
A decimal number is a fraction whose denominator is 10 or some multiple of 10.
Decimal Point
A decimal point is the place marker in a decimal number that separates the whole number and the fraction part. The decimal number 326.45 has the decimal point between the six and the four.
Magnitude
A magnitude is the size of a number without respect to its direction. The number -35.6 has a direction to the left and a magnitude of 35.6.

Guided Practice

1. Subtract these decimal numbers: \begin{align*}(243.67)-(196.3579)\end{align*}
2. \begin{align*}(32.47)-(-28.8)-(19.645)\end{align*}
3. Josie has $59.27 in her bank account. She went to the grocery store and wrote a cheque for$62.18 to pay for the groceries. Describe Josie’s balance in her bank account now.

1. \begin{align*}(243.67)-(196.3579)\end{align*}

Write the decimal numbers using the vertical alignment method.

\begin{align*}& \quad 243.67\\ & \underline{-196.3579}\\ & \end{align*}

To ensure that the digits are properly aligned, add zeros to 243.67

\begin{align*}& \quad 243.67 {\color{blue}00}\\ & \underline{-196.3579}\\ & \end{align*}

Subtract the numbers. Work from right to left and borrow when it is necessary.

\begin{align*}& \ \ \overset{1}{\cancel{2}} \ \overset{13}{\cancel{4}} \ {^1} 3.6 \overset{6}{\cancel{7}} \ \overset{9}{^1 {\color{blue}\bcancel{0}}} \ {^1} {\color{blue}0}\\ & \underline{ -1 \; 9 \; \;\; 6.35 \;\; 7 \;\;\; 9\;\;}\\ & \quad \ \ 4 \ \ 7.31 \ \ 2 \ \ 1\end{align*}

2. \begin{align*}(32.47)-(-28.8)-(19.645)\end{align*}

Write the question as an addition problem and change the sign of the original number being subtracted.

\begin{align*}(32.47)+(+28.8)+(-19.645)\end{align*}

\begin{align*}& \ \ 32.47 \qquad \text{Add a zero to} \ 28.8\\ & \underline{ +28.8 {\color{blue}0}}\\ & \end{align*}

\begin{align*}& \ \ 32.47\\ & \underline{ +28.8 {\color{blue}0}}\\ & \ \ 61.27\end{align*}

The numbers being added are both positive so the answer will also be positive.

\begin{align*}(+61.27)+(-19.645)\end{align*}

Write the problem using the vertical alignment method.

\begin{align*}& \quad 61.27 {\color{blue}0}\\ & \underline{-19.645}\\ & \end{align*}

The number of greater magnitude was written above the smaller number. A zero was added to 61.27. The numbers have opposite signs so they will be subtracted and the answer will have the same sign as the larger number – positive.

\begin{align*}& \quad 61.27 {\color{blue}0}\\ & \underline{-19.645}\\ & \ \ 41.625\end{align*}

3. Amount in her bank account - $59.27 Amount of the written cheque -$62.18

The amount of the cheque is greater than the amount of money in the account.

The account will have a negative value. This means that her account is overdrawn.

Summary

The subtraction of decimal numbers is simply the subtraction of whole numbers and like fraction parts. To make this process simpler, the decimal numbers are written using the vertical alignment method with the number of greater magnitude being written above the number of lesser magnitude. The decimal points are aligned as well as the numbers are aligned according to their place value. The numbers in each vertical column are then subtracted by starting right and working to the left. If the decimal numbers are signed numbers, the rules for subtracting integers are applied to the problem.

Problem Set

Subtract the following decimal numbers:

1. \begin{align*}42.37-15.32\end{align*}
2. \begin{align*}37.891-7.2827\end{align*}
3. \begin{align*}579.237-45.68\end{align*}
4. \begin{align*}4.2935-0.327\end{align*}
5. \begin{align*}16.074-7.58\end{align*}

Subtract the following signed decimal numbers:

1. \begin{align*}(-17.39)-(-49.68)\end{align*}
2. \begin{align*}(92.75)+(-106.682)\end{align*}
3. \begin{align*}(-72.5)-(-77.57)-(31.724)\end{align*}
4. \begin{align*}(-82.456)-(279.83)+(-567.3)\end{align*}
5. \begin{align*}(-57.76)-(-85.9)-(33.84)\end{align*}

Determine the answer to the following problems.

1. The diameter of No. 12 bare copper wire is 0.08081 in., and the diameter of No. 15 bare copper wire is 0.05707 in. How much larger is No.12 wire than No. 15 wire?
2. The resistance of an armature while it is cold is 0.208 ohm. After running for several minutes, the resistance increases to 1.340 ohms. Find the increase in resistance of the armature.
3. The highest temperature recorded in Canada this year was \begin{align*}114.8^\circ F\end{align*}. The lowest temperature of \begin{align*}-62.9^\circ F\end{align*} was recorded in February this year. Find the difference between the highest and lowest temperatures recorded in Canada this year.
4. The temperature in Alaska was recorded as \begin{align*}-78.64^\circ F\end{align*} in January of 2010 and as \begin{align*}-59.8^\circ F\end{align*} on the same date in 2011. What is the difference between the two recorded temperatures?
5. Laurie has a balance of -32.16 in her bank account. Write a problem that could represent this balance. Answers Subtract the following decimal numbers: 1. \begin{align*}42.37-15.32\end{align*} Use the vertical alignment method. \begin{align*}& \quad \overset{3}{\cancel{4}} \ {^1} 2.37\\ & \underline{ \; -1 \; \; 5.32}\\ & \quad 2 \ \ 7.05\end{align*} Subtract right to left and borrow when it is necessary. 1. \begin{align*}579.237-45.68\end{align*} Use the vertical alignment method. \begin{align*}& \ 57 \overset{8}{\cancel{9}}.\overset{11}{\cancel{2}} \ ^1 37\\ & \underline{-45. 6 \quad 8{\color{blue}0}}\\ & \ 533. 5 \quad 57\end{align*} Add a zero to 45.68 and subtract the numbers. 1. \begin{align*}16.074-7.58\end{align*} Use the vertical alignment method. \begin{align*}& \overset{0}{\cancel{1}} \ \overset{15}{\cancel{6}}. \ \overset{9}{^1\bcancel{0}} \ {^1} 74\\ & \underline{ - \; 7. \; \; 5 \; \; 8{\color{blue}0}}\\ & \quad \ 8. \ 4 \ \ 94\end{align*} Add a zero to 7.58 and subtract the numbers. Subtract the following signed decimal numbers: \begin{align*}&(-17.39)-(-49.68)\\ &(-17.39)+(+49.68)\end{align*} Write the problem as an addition problem and change the sign of the original number being subtracted. \begin{align*}& \quad 49.68\\ & \underline{ - 17.39\;}\end{align*} Write the problem using the vertical alignment method. Place the number of greater magnitude above the number of lesser magnitude. Follow the rules for adding integers. \begin{align*}& \quad 49. \overset{5}{\cancel{6}} \ {^1} 8\\ & \underline{ - 17. 3 \quad 9}\\ & \quad 32.2 \ \ 9\end{align*} 1. \begin{align*}(-72.5)-(-77.57)-(31.724)\end{align*} Write the problem as an addition problem and change the sign of the original numbers being subtracted. \begin{align*}(-72.5)+(+77.57)+(-31.724)\end{align*} \begin{align*}& \ \ -72.5 {\color{blue}00}\\ & \underline{ \quad -31. 724}\\ & -104.224\end{align*} Write the problem using the vertical alignment method. Add zeros to -72.5 and follow the rules for the addition of integers. Follow the rules for adding integers. Numbers with the same sign must be added and the sign of the numbers being added will be the sign of the answer. \begin{align*}(-104.224)+(+77.57)\end{align*} The numbers being added have opposite signs. Place the number of greater magnitude on top. Subtract the numbers and apply the sign of the larger number to the answer. \begin{align*}& - \overset{9}{\bcancel{10}} \ \overset{13}{\bcancel{4}}. \overset{11}{\bcancel{2}} \ {^1} 24\\ & \ \underline{+ \; \;\; 7 \; 7. \;5 \;\;\; 7{\color{blue}0}}\\ & - \ \ 2 \ 6. \; 6 \quad 54\end{align*} \begin{align*}& (-57.76)-(-85.9)-(33.84)\\ & (-57.76)+(+85.9)+(-33.84)\end{align*} \begin{align*}& -57.76 \quad \ \ -91.60\\ & \underline{ \; -33.84} \qquad \underline{ \; +85.90}\\ & -91.60 \qquad - 5.70\end{align*} Determine the answer to the following problems. 1. \begin{align*}0.08081-0.05707\end{align*} \begin{align*}& \quad 0.0 \ \overset{7}{\bcancel{8}} \ {^1} 0 \ \overset{7}{\bcancel{8}} \ {^1} 1\\ & \underline{ \; -0.0 \; 5 \; \; 7 \; 0 \; \;\; 7\;\;}\\ & \quad 0.0 \ 2 \ \ 3 \ 7 \ \ 4\end{align*} The No. 12 wire is 0.02374 in. thicker than No. 15 copper wire. \begin{align*}& 114.8-(-62.9)\\ & 114.8+(+62.9)\end{align*} \begin{align*}& \ 114.8\\ & \underline{+62.9}\\ & \ 177.7\end{align*} The difference between the lowest and highest recorded temperatures is \begin{align*}177.7^\circ F\end{align*}. 1. -32.16 is the balance of the bank account. Laurie had a balance of $243.39 in her bank account. Her car payment of$275.55 was taken from the same account. This transaction left Laurie with a negative balance of \$32.16.

## Summary

In this lesson you have learned how to subtract real numbers by using a variety of models. The real numbers that you subtracted were integers, fractions and decimal numbers. The models that were used to subtract integers were color counters, algebra tiles and a number line. By using these models, you learned that subtracting integers was actually adding the opposite. The subtraction problems were written as addition problems and the sign of the original number being subtracted was changed. To calculate the answer, the two rules for adding integers were applied. After subtracting integers, you then learned how to subtract fractions by using fraction strips and a number line. You learned that fractions can only be subtracted if they have a common denominator. The difference of the numerators of the fractions being subtracted is placed over the common denominator. The last real numbers that were subtracted were decimal numbers. You learned that a number has both direction and magnitude. The direction of a positive number is to the right and that of a negative number is to the left. This direction is determined by the numbers location with respect to zero on the number line. The magnitude of a number is simply its size with no regard to its sign. When subtracting decimal numbers that were signed numbers, the rules for the addition of integers were applied to the problems.

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