Skip Navigation

Subtraction of Rational Numbers

Subtracting fractions

Atoms Practice
Estimated9 minsto complete
Practice Subtraction of Rational Numbers
This indicates how strong in your memory this concept is
Estimated9 minsto complete
Practice Now
Turn In
Subtraction of Rational Numbers


Tara is hiking down a hill. Starting from a point 20 feet due west of the peak and 20 feet down, she climbs down to a point 30 feet due west of the peak, and 40 feet down. If the slope of the hill is described as the ratio of the vertical travel to the horizontal travel, how steep is the hill?

Subtraction of Rational Numbers

Suppose you want to find the difference of 9 and 12. Symbolically, this would be 9 - 12. Begin by placing a dot at 9 and move to the left 12 units.

\begin{align*}9 - 12 = \text{-}3\end{align*}

To subtract a number, add its opposite. Here are a couple of examples:

\begin{align*}3 - 5 = 3 + (\text{-}5) = \text{-}2 && 9 - 16 = 9 + (\text{-}16) = \text{-}7\end{align*}

A special case of this rule can be used to subtract a negative number.

The Opposite-Opposite Property states that for any real numbers \begin{align*}a \text{ and }b:\end{align*} \begin{align*}a-(\text{-}b) = a + b\end{align*}

Now, apply the information from above and simplify the following expressions:

  1. \begin{align*}\text{-}6 - (\text{-}13)\end{align*} 

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

\begin{align*}\text{-}6 - (\text{-}13) = \text{-}6 + 13 = 7\end{align*}

  1. \begin{align*}\frac{5}{6} - \left ( \text{-} \frac{1}{18} \right )\end{align*}

Begin by using the Opposite-Opposite Property.

\begin{align*}\frac{5}{6} + \frac{1}{18}\end{align*}

Next, create a common denominator:

\begin{align*}\frac{5 \times 3}{6 \times 3} + \frac{1}{18} = \frac{15}{18} + \frac{1}{18}\end{align*}

Add the fractions:

\begin{align*}\frac{15}{18} + \frac{1}{18} = \frac{16}{18}\end{align*}

Finally, reduce:

\begin{align*}\frac{2 \times 2 \times 2 \times \cancel{2}}{3 \times 3 \times \cancel{2}} = \frac{8}{9}\end{align*}

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. This will tell you what the average change is in the dependent variable compared to the independent variable. 

Consider this table of the number of song downloads purchased compared to the total cost:

\begin{align*}&\text{Number of songs} && 2 && 4 && \hphantom{1}6 && \hphantom{1}8 && 10\\ &\text{Cost (\$)} && 4 && 8 && 12 && 16 && 20\end{align*}

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

Begin by finding the difference between the cost of any two values. For example, the change in cost between 4 songs and 8 songs.


Next, find the difference between the number of songs:


Finally, divide the change in cost by the change in number of songs:

\begin{align*}\frac{8}{4}=2\end{align*}Each song costs an average of $2.


Example 1

Earlier, you were asked about how you would find the steepness of a hill that has a point at 20 feet over and 20 feet down from the peak (20, -20), and another point at 30 feet over and 40 feet down (30, -40). The steepness is the ratio of the difference between the vertical values (the dependent variable, \begin{align*}y\end{align*}) divided by the difference of the horizontal values (the independent variable, \begin{align*}x\end{align*}). As a formula, this looks like: \begin{align*}\text{slope}= \tfrac{rise}{run}=\tfrac{y_2 - y_1}{x_2 - x_1}.\end{align*}

The independent variable is the first coordinate, the \begin{align*}x\end{align*}-coordinate.  The difference in independent values is: \begin{align*}30-20=10\end{align*}

The dependent variable is the second coordinate, the \begin{align*}y\end{align*}-coordinate. The difference in dependent values is: \begin{align*}\text{-}40-(\text{-}20)=\text{-}40+20=\text{-}20.\end{align*} (Remember to use the Opposite-Opposite property as needed.)

Now, dividing the difference in dependent values by the difference in independent values, we get that the steepness is \begin{align*}\frac{\text{-}20}{10}=\frac{\text{-}2}{1} = \text{-}2.\end{align*}

The slope of the hill is -2, that means that, for every 1 foot Tara travels away from the peak horizontally, she descends 2 feet vertically. That is a pretty steep hill!

Example 2

For the equation \begin{align*}y=3x-5\end{align*}, find the difference between the dependent variable \begin{align*}(y)\end{align*} values when \begin{align*}x=2\end{align*} and when \begin{align*}x=4\end{align*}. Then, find the steepness of the line.

First, substitute each given \begin{align*}x\end{align*} value into the equation, to find the corresponding \begin{align*}y\end{align*} values:

\begin{align*}\text{When }x \text{ is }2: \ y=3x-5=3(2)-5=6-5=1\end{align*}

\begin{align*}\text{When }x \text{ is }4: \ y=3x-5=3(4)-5=12-5=7\end{align*}

Now, we calculate the difference in the dependent \begin{align*}(y)\end{align*} values:


Finally, to find the steepness (also called slope) we divide the difference in \begin{align*}y\end{align*} values by the difference between the independent \begin{align*}(x)\end{align*} values:

\begin{align*}\text{Difference between the }x \text{ values}: \ 4-2=2\end{align*}

\begin{align*}\frac{\text{Difference in } y \text{ values}}{\text{Difference in }x \text{ values}}=\frac{6}{2}=3\end{align*}

The steepness is 3. Notice that this is the value in front of \begin{align*}x\end{align*} in the equation, that will be an important point when you study the slope of a line.


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

  1. \begin{align*}9 - 14\end{align*}
  2. \begin{align*}2 - 7\end{align*}
  3. \begin{align*}21 - 8\end{align*}
  4. \begin{align*}8 - (\text{-}14)\end{align*}
  5. \begin{align*}\text{-}11 - (\text{-}50)\end{align*}
  6. \begin{align*}\vphantom{\Bigg[} \frac{5}{12} - \frac{9}{18}\end{align*}
  7. \begin{align*}5.4 - 1.01\end{align*}
  8. \begin{align*}\vphantom{\Bigg[} \frac{2}{3} - \frac{1}{4}\end{align*}
  9. \begin{align*}\vphantom{\Bigg[} \frac{3}{4} - \frac{1}{3}\end{align*}
  10. \begin{align*}\vphantom{\Bigg[} \frac{1}{4} - \left (\text{-} \frac{2}{3} \right )\end{align*}
  11. \begin{align*}\vphantom{\Bigg[} \frac{15}{11} - \frac{9}{7}\end{align*}
  12. \begin{align*}\vphantom{\Bigg[} \frac{2}{13} - \frac{1}{11}\end{align*}
  13. \begin{align*}\vphantom{\Bigg[} \text{-}\frac{7}{8} - \left (\text{-} \frac{8}{3} \right )\end{align*}
  14. \begin{align*}\vphantom{\Bigg[} \frac{7}{27} - \frac{9}{39}\end{align*}
  15. \begin{align*}\vphantom{\Bigg[} \frac{6}{11} - \frac{3}{22}\end{align*}
  16. \begin{align*}\vphantom{\Bigg[} \text{-}3.1 - 21.49\end{align*}
  17. \begin{align*}\vphantom{\Bigg[} \frac{13}{64} - \frac{7}{40}\end{align*}
  18. \begin{align*}\vphantom{\Bigg[} \frac{11}{70} - \frac{11}{30}\end{align*}
  19. \begin{align*}\vphantom{\Bigg[} \text{-}68 - (\text{-}22)\end{align*}
  20. \begin{align*}\vphantom{\Bigg[} \frac{1}{3} - \frac{1}{2}\end{align*}
  21. Determine the steepness of the line between (1, 9) and (5, –14).
  22. Consider the equation \begin{align*}y = 3x + 2.\end{align*} Determine the steepness in the line between \begin{align*}x = 3\end{align*} and \begin{align*}x = 7.\end{align*}
  23. Consider the equation \begin{align*}y = \tfrac{2}{3}x + \tfrac{1}{2}.\end{align*} Determine the steepness in the line between \begin{align*}x = 1\end{align*} and \begin{align*}x = 2.\end{align*}
  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 \begin{align*}a=2, \ b= \text{-}3, \text{ and }c=\text{-}1.5.\end{align*}

  1. \begin{align*}(a-b)+c\end{align*}
  2. \begin{align*}|b+c|- a\end{align*}
  3. \begin{align*}a-(b+c)\end{align*}
  4. \begin{align*}|b|+ |c|+ a\end{align*}
  5. \begin{align*}7b+4a\end{align*}
  6. \begin{align*}(c-a)- b\end{align*}

Review (Answers)

To see the Review answers, open this PDF file and look for section 2.5. 

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More


dependent variable A dependent variable is one whose values depend upon what is substituted for the other variable.
difference Subtraction.
Opposite-Opposite Property For any real numbers a and b, \ a-(-b) = a + b.
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.

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Subtraction of Rational Numbers.
Please wait...
Please wait...