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# 2.3: Subtraction of Rational Numbers

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

At the end of this lesson, students will be able to:

• Subtract rational numbers.
• Evaluate change using a variable expression.
• Solve real world problems using fractions.

## Vocabulary

Terms introduced in this lesson:

opposite
opposite/inverse process
subtraction
method of prime factors
simplify
simplest form
invisible denominator
speed
difference
change

## Teaching Strategies and Tips

General Tip: Additive inverses provide a way to subtract two real numbers. Students learn that the operation of subtraction can be replaced with addition by adding the additive inverse of the number.

Example:

\begin{align*}\frac{3} {7} - \frac{6} {7} = \frac{3} {7} + \left (-\frac{6} {7}\right ) = \frac{3 + (-6)} {7} = \frac{-3} {7}\end{align*}

Although it may be easier for students to keep it as subtraction:

\begin{align*}\frac{3} {7} - \frac{6} {7} = \frac{3 - 6} {7} = \frac{3 + (-6)} {7} = \frac{-3} {7}\end{align*}

Tips on factor trees:

• Do not include the trivial factors of \begin{align*}1\end{align*} in the branches of factor trees.
• The first two branches of the factor tree can be any pair of factors.
• For instance, in Example 2, \begin{align*}90 = 9 \times 10\end{align*} and \begin{align*}126 = 9 \times 14\end{align*}. Students may choose differently: \begin{align*}90 = 45 \times 2\end{align*} and \begin{align*}126 = 21 \times 6\end{align*}.
• To find the LCM of three numbers, no more than three factor trees need to be constructed. In general, there will be as many factor trees as numbers whose LCM is being sought. See Example 4 for three numbers.

In Example 4, the “invisible denominator” is a useful way of getting students to realize that

\begin{align*}x = \frac{x} {1}\end{align*}

That is, integers are rational numbers. Every whole number can be viewed as a rational number whose denominator is \begin{align*}1\end{align*}.

Use Examples 5 and 6 to demonstrate the concept of change.

• A variable expression is used to evaluate change in speed and change in light intensity, respectively. The functions are provided already; and therefore do not need to be derived from scratch.
• The focus is on the concept of change, which will be new for many students. Exploration is encouraged through tables and graphs.
• To find the change in a quantity using a function, cast this as a subtraction problem: subtract first quantity from second quantity.
• Explaining the results in words will benefit students. Positive change in speed means an increase in speed. The change is negative in Example 6: the intensity reduces as the train travels farther away from the light source.

## Error Troubleshooting

General Tip: For many students, the combination of subtraction and finding common denominators is prone to errors. Practice is crucial. See Examples 3 and 4 as well as Review Questions 1a – 1i.

Example 5 and 6: \begin{align*}\mathrm{Change} = \mathrm{second\ quantity} - \;\mathrm{first\ quantity}\end{align*}; i.e., subtract the first quantity from the second quantity. In context:

• \begin{align*}\mathrm{Change} = \mathrm{Speed}\ 2 - \mathrm{Speed}\ 1\end{align*}
• \begin{align*}\mathrm{Change} = \mathrm{Intensity}\ 2 - \mathrm{Intensity}\ 1\end{align*}

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Feb 22, 2012
Aug 22, 2014
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CK.MAT.ENG.TE.1.Algebra-I.2.3