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2.5: Subtraction of Rational Numbers

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Suppose that you know that two points on a line are (2, -4) and (3, -2) . How would you go about finding the "steepness" of the line? As you'll learn in a future concept, it is the difference of the dependent variable divided by the difference of the independent variable. Do you know how to do this? After completing this Concept, you will.

Guidance

Subtraction of Rational Numbers

In the previous two concepts, you have learned how to find the opposite of a rational number and them together. You can use these two ideas to subtract rational numbers as well. Suppose you want to find the difference of 9 and 12. Symbolically, it would be 9 - 12 . Begin by placing a dot at nine and move to the left 12 units.

9 - 12 = -3

Rule: To subtract a number, add its opposite.

3 - 5 = 3 + (-5) = -2 && 9 - 16 = 9 + (-16) = -7

A special case of this rule can be written when trying to subtract a negative number.

The Opposite-Opposite Property: For any real numbers a and b, \ a-(-b) = a + b .

Example A

Simplify -6 - (-13).

Solution: Using the Opposite-Opposite Property, the double negative is rewritten as a positive.

-6 - (-13) = -6 + 13 = 7

Example B

Simplify \frac{5}{6} - \left ( - \frac{1}{18} \right ).

Solution: Begin by using the Opposite-Opposite Property.

\frac{5}{6} + \frac{1}{18}

Next, create a common denominator: \frac{5 \times 3}{6 \times 3} + \frac{1}{18} = \frac{15}{18} + \frac{1}{18}.

Add the fractions: \frac{15}{18} + \frac{1}{18} = \frac{16}{18}.

Reduce: \frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 2} = \frac{8}{9}.

Evaluating Change Using a Variable Expression

You have learned how to graph a function by using an algebraic expression to generate a table of values. Using the table of values you can find the difference of the dependent values relative to the difference of the two independent values.

Example C

In a previous concept, you wrote an expression to represent the pattern of the total cost to the number of CDs purchased. The table is repeated below:

&\text{Number of CDs} && 2 && 4 && 6 && 8 && 10\\&\text{Cost (\$)} && 24 && 48 && 72 && 96 && 120

To determine the average cost of the CDs, you must find the difference between the dependent values and divide it by the difference in the independent values.

Solution: We begin by finding the difference between the cost of two values. For example, the change in cost between 4 CDs and 8 CDs.

96-48 = 48

Next, we find the difference between the number of CDs.

&&&8 - 4  = 4\\&\text{Finally, we divide.}&& \quad \frac{48}{4}=12

This tells us that the cost for each CD was 12 dollars.

Guided Practice

For the equation y=3x-5 , find the difference between the dependent variable from x=2 and x=4 . Then, find the steepness of the line.

Solution: First we find the value of the equation, or the value of y for each x value:

y=3x-5=3(2)-5=6-5=1

y=3x-5=3(4)-5=12-5=7

Now, we calculate the difference in the dependent values:

7-1=6

Finally, to find the steepness (also called slope in later concepts) we divide by the difference between the independent values:

4-2=2

\frac{6}{2}=3

The steepness is 3. Notice that this is the value in front of x in the equation. We will learn more about that in future concepts.

Practice

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Subtraction of Rational Numbers (10:22)

In 1 – 20, subtract the following rational numbers. Be sure that your answer is in the simplest form.

  1. 9 - 14
  2. 2 - 7
  3. 21 - 8
  4. 8 - (-14)
  5. -11 - (-50)
  6. \frac{5}{12} - \frac{9}{18}
  7. 5.4 - 1.01
  8. \frac{2}{3} - \frac{1}{4}
  9. \frac{3}{4} - \frac{1}{3}
  10. \frac{1}{4} - \left (- \frac{2}{3} \right )
  11. \frac{15}{11} - \frac{9}{7}
  12. \frac{2}{13} - \frac{1}{11}
  13. -\frac{7}{8} - \left (- \frac{8}{3} \right )
  14. \frac{7}{27} - \frac{9}{39}
  15. \frac{6}{11} - \frac{3}{22}
  16. -3.1 - 21.49
  17. \frac{13}{64} - \frac{7}{40}
  18. \frac{11}{70} - \frac{11}{30}
  19. -68 - (-22)
  20. \frac{1}{3} - \frac{1}{2}
  21. Determine the steepness of the line between (1, 9) and (5, –14).
  22. Consider the equation y = 3x + 2 . Determine the steepness in the line between x = 3 and x = 7.
  23. Consider the equation y = \frac{2}{3}x + \frac{1}{2} . Determine the steepness in the line between x = 1 and x = 2 .
  24. True or false? If the statement is false, explain your reasoning. The difference of two numbers is less than each number.
  25. True or false? If the statement is false, explain your reasoning. A number minus its opposite is twice the number.
  26. KMN stock began the day with a price of $4.83 per share. At the closing bell, the price dropped $0.97 per share. What was the closing price of KMN stock?

In 27 – 32, evaluate the expression. Assume a=2, \ b= -3, and c = -1.5.

  1. (a-b)+c
  2. |b+c|- a
  3. a-(b+c)
  4. |b|+ |c|+ a
  5. 7b+4a
  6. (c-a)- b

Mixed Review

  1. Graph the following ordered pairs: \left \{(0,0),(4,4),(7,1),(3,8) \right \} . Is the relation a function?
  2. Evaluate the following expression when m= \left (- \frac{2}{3} \right ): \ \frac{2^3+m}{4} .
  3. Translate the following into an algebraic equation: Ricky has twelve more dollars than Stacy. Stacy has 5 less dollars than Aaron. The total of the friends’ money is $62.
  4. Simplify \frac{1}{3} + \frac{7}{5} .
  5. Simplify \frac{21}{4} - \frac{2}{3} .

Vocabulary

dependent variable

dependent variable

A dependent variable is one whose values depend upon what is substituted for the other variable.
difference

difference

Subtraction.
Opposite-Opposite Property

Opposite-Opposite Property

For any real numbers a and b, \ a-(-b) = a + b.
steepness (of a line)

steepness (of a line)

The difference between the dependent variables divided by the difference of the independent variables. This is also called the slope of a line. It can also be referred to as the change in the dependent variables.

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Difficulty Level:

Basic

Grades:

8 , 9

Date Created:

Feb 24, 2012

Last Modified:

Aug 21, 2014
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