To add or subtract algebraic fractions with different denominators, we’ll use – just as you’d expect – the “multiply by 1 trick” to give the fractions a common denominator. For example,
Example 2. Add and simplify: $ + $ .
Solution. For a common denominator, we need a multiple of $ 2a^{2}b $ and 6ab. The “smallest” such denominator is $ 6a^{2}b $ . Using the “multiply by 1 trick,” we’ll convert to this new denominator.
\[ \begin{aligned}\frac{3}{2a^{2}b}+\frac{7}{6ab}&=\left(\frac{3}{2a^{2}b}\cdot\frac{3}{3}\right)+\left(\frac{7}{6ab}\cdot\frac{a}{a}\right)\quad&(the“multiply by1trick”)\\&=\frac{9}{6a^{2}b}+\frac{7a}{6a^{2}b}\\&=\frac{9+7a}{6a^{2}b}\end{aligned} \]
After you’ve become comfortable adding and subtracting fractions with different denominators, you need not write out every step. In practice, we usually condense the process as follows:
Addition and Subtraction Rule for Fractions. We find the denominator (a so-called “common denominator”) as follows: Take any multiple of the given denominators.* We find the numerator as follows: Multiply each given numerator by the factor that will turn its denominator into the common denominator. These products will be the terms in the new numerator.
If we re-do the previous example using this shortcut addition rule, we’ll have somewhat less to write. Taking $ 6a^{2}b $ as our common denominator, the addition rule quickly tells us that
\[ \frac{3}{2a^{2}b}+\frac{7}{6ab}=\frac{3(3)+7(\boldsymbol{a})}{6a^{2}b}=\frac{9+7a}{6a^{2}b}. \]
Beginning students frequently make mistakes when subtracting fractions – usually because they omit necessary parentheses. Be especially careful when subtracting.
Example 3. Subtract and simplify: $ - $ .
\[ \begin{aligned}Solution.\ \frac{3}{x-5}-\frac{2x-1}{x+5}&=\frac{[3(x+5)]-[2x-1)(x-5)]}{(x-5)(x+5)}\quad&(subtraction rule)\\&=\frac{[3x+15]-[2x^{2}-11x+5]}{(x-5)(x+5)}\quad&(distributing within the brackets)\\&=\frac{3x+15-2x^{2}+11x-5}{x^{2}-25}\quad&(distributing a minus;difference of squares)\\&=\frac{-2x^{2}+14x+10}{x^{2}-25}\quad&(combining like terms)\quad\diamond\end{aligned} \]
Exercises
Carefully explain why $ + = $ .
Simplify the following expressions.
$ (+) $
$ --(-) $
$ 1 - $
In Example 2 above, we took $ 6a^{2}b $ as our common denominator. Suppose we had used $ 12a{3}b{2} $ instead. Would this have changed the result? Work it out that way and see.
Express each of the following as a single fraction, and simplify as much as possible:
$ + $ b) $ ()() $ c) $ - $ d) $ + $
$ + $ f) $ - $ g) $ + + $ h) $ - + - $
$ - $
$ - $
$ 3 - $
$ 1 - $
$ - $ n) $ - + - $
Last Words and One Nasty Example
You can now add, subtract, multiply, or divide any two algebraic fractions. No lingering mysteries remain; you are fully initiated. Should you ever need to combine seventeen fractions linked in all sorts of intricate arithmetical ways, you have all the knowledge necessary to do it. All you must do is slow down and take each piece in order. It may be tedious, but it shouldn’t be difficult.
To illustrate my point, I’ll offer one last example. If you have understood everything so far in this chapter, it should pose no conceptual difficulties, even though it is rather involved.
Problem. Simplify the following ugly expression as much as possible: $ + $
Solution. Before we begin hacking away at this overgrown mess, let’s consider it from a distance to convince ourselves that, despite its ugliness, this is something we can handle. Begin by observing that the expression has two terms. True, the first one looks nasty (“Ugh! Fractions in a fraction!”), until we realize that when we subtract the two fractions in its numerator, they’ll obviously combine into a single fraction, which we can then divide by the fraction in the left term’s denominator. The result of that division will obviously be… a single fraction. Thus, when the dust settles, we’ll have rewritten the whole ugly first term as one fraction. All the “fractions within fractions” will be gone. As for the second term, you can probably simplify it in your head, turning it into one fraction, too. Then, all that will remain is to add two ordinary algebraic fractions – an easy task.
Having watched the preliminary dumbshow, let’s proceed to the actual details of calculation, confident that, at least in outline, we already know how matters will work out.
The first term’s numerator is
\[ \begin{aligned}\frac{2}{x-2}-\frac{x+1}{x+2}&=\frac{\left[2(x+2)\right]-\left[(x-2)(x+1)\right]}{(x-2)(x+2)}\\&=\frac{\left[2(x+2)\right]-\left[(x-2)(x+1)\right]}{x^{2}-4}\\&=\frac{\left[2x+4\right]-\left[x^{2}-x-2\right]}{x^{2}-4}\\&=\frac{2x+4-x^{2}+x+2}{x^{2}-4}\\&=\frac{-x^{2}+3x+6}{x^{2}-4}\end{aligned} \]
(note the brackets!)
(difference of squares)
(distributing within the brackets)
(removing brackets, distributing the negative)
(combining like terms).
Now that we’ve simplified the first term’s numerator, we’ll divide it by the first term’s denominator. Doing so, we find that the entire first term simplifies to
\[ \begin{aligned}\frac{\frac{-x^{2}+3x+6}{x^{2}-4}}{\frac{3x-7}{x^{2}-4}}&=\frac{-x^{2}+3x+6}{x^{2}-4}\cdot\frac{x^{2}-4}{3x-7}\quad&(division rule for fractions)\\&=\frac{\left(-x^{2}+3x+6\right)\left(x^{2}-4\right)}{(x^{2}-4)(3x-7)}\quad&(multiplication rule for fractions)\\&=\frac{-x^{2}+3x+6}{3x-7}\quad&(cancelling(x^{2}-4)above and below the bar).\end{aligned} \]
Having reduced the first term to something manageable, we can tackle the original problem:
\[ \begin{aligned}\frac{\frac{2}{x-2}-\frac{x+1}{x+2}}{\frac{3x-7}{x^{2}-4}}+\frac{\frac{2}{x^{2}}}{-5}&=\frac{-x^{2}+3x+6}{3x-7}+\frac{\frac{2}{x^{2}}}{-5}\\&=\frac{-x^{2}+3x+6}{3x-7}-\frac{2}{5x^{2}}\\&=\frac{\left[(-x^{2}+3x+6)(5x^{2})\right]-\left[2(3x-7)\right]}{(3x-7)(5x^{2})}\\&=\frac{\left[-5x^{4}+15x^{3}+30x^{2}\right]-\left[6x-14\right]}{15x^{3}-35x^{2}}\\&=\frac{-5x^{4}+15x^{3}+30x^{2}-6x+14}{15x^{3}-35x^{2}}.\end{aligned} \]
(by our work on the first term above)
(division of fractions, second term)
(subtraction rule for fractions)
(a great flurry of multiplication)
(removing brackets upstairs)
There were many steps in that last problem, but each was simple. Occasional mistakes in such problems are inevitable, but bear in mind that there are mistakes and mistakes. Accidentally writing $ 3 = 6 $ in the midst of a larger problem is wrong, but it presumable isn’t a conceptual error. On the other hand, omitting crucial parentheses, forgetting to distribute negatives, or botching the multiply by 1 trick are serious mistakes that most likely stem from conceptual misunderstandings. If you intend to take further mathematics courses, you need to clear up any and all such misunderstandings immediately. Those courses will give you plenty of new material to think about; if you are still struggling with basic algebra at that stage, you won’t see the forest for the trees.
Exercises
- Simplify as much as possible:
- $ $ b) $ $ c) $ ()(b+c) $ d) $ ()b+c $
- True or false (explain your answers):
$ -3^{2} = 9 $
-x always represents a negative number.
$ (-3)^{2} = -9 $
$ a - b = -(b - a) $ .
$ = $
$ = $
$ = $
- Express as a single fraction – and simplify as much as possible:
$ + $ [Hint: You may find Exercise 44d useful.]
$ $
$ $
$ $
\[ \mathrm{e})\left[\frac{5x+4}{x+1}-\frac{-3x^{2}+9x+4}{(x+1)^{2}}\right]\div\left(\frac{4x^{3}}{(x+1)^{2}}\right) \]
$ 1 + $
$ $
$ $
$ $
\[ \mathrm{j})\left(\frac{x^{2}}{1+x^{2}}\cdot\frac{\left(1+x^{2}\right)\left(-x^{2}\right)}{x^{4}}\right)(a-b) \]
- Early in the chapter, I claimed that all of fractional arithmetic follows logically from “two intuitive rules”:
\[ \frac{a}{b}=a\left(\frac{1}{b}\right)\text{and}\left(\frac{1}{a}\right)\left(\frac{1}{b}\right)=\frac{1}{ab}. \]
Now that you know all the rules of fractional arithmetic, it’s worth reexamining my claim. I explicitly pointed out in the text how the multiplication rule for fractions follows directly from the two intuitive rules. What about cancelling factors above and below the bar? Well, we saw that this operation is really just a matter of detaching a hidden factor of 1, like so:
\[ \frac{6ab}{3bc}=\frac{3b\cdot2a}{3b\cdot c}=\left(\frac{3b}{3b}\right)\left(\frac{2a}{c}\right)=1\left(\frac{2a}{c}\right)=\frac{2a}{c}. \]
The “detachment” (at the second equals sign) justified by the multiplication rule… which was built on the two intuitive rules. (The other steps are justified by completely obvious facts such as “multiplying by 1 doesn’t change anything” or “anything divided by itself is 1”.) Thus cancellation above and below the bar is ultimately a logical consequence of the two intuitive rules.
Convince yourself that the “multiply by 1 trick” is ultimately justified by the two intuitive rules.
Do the same for the division rule for fractions.
Do the same for the addition and subtraction rule for fractions.
Congratulate yourself: You’ve now seen that the whole vexed subject of fractions is actually quite simple, built up from just a few intuitive rules and the logical consequences thereof.