Adding and subtracting rational expressions
Two fractions have polynomials in their denominators. What denominator can they share, so that they become one fraction?
Where you are 3 in Simplifying rational expressions
When you reach for it
Rates, costs and formulas often arrive as fractions with a variable in the denominator. If Jacoba squeezes a jug of lemonade in t minutes, she squeezes \displaystyle \frac{1}{t} of a jug each minute, and the work of two people together is a sum of two such fractions. One fraction is easier to read, to evaluate, to graph and to set equal to something than a sum of several.
The arithmetic is the arithmetic of number fractions from adding and subtracting fractions: rename both over a common denominator, then combine the numerators. Two things are new. The common denominator is now a polynomial, so it has to be built from factors. And a fraction can sit inside another fraction, which is called a complex fraction.
The lesson runs over two days. Part 1 adds and subtracts rational expressions. Part 2 simplifies complex fractions. Both lean on simplifying a rational expression and its excluded values and on multiplying and dividing rational expressions.
The procedure
Part 1: sums and differences
With a common denominator. For polynomials P, Q and R,
for every value of the variable where R \ne 0. The reason is the distributive property: \displaystyle \frac{P}{R} is \displaystyle P \cdot \frac{1}{R}, and \displaystyle P \cdot \frac{1}{R} + Q \cdot \frac{1}{R} = (P + Q) \cdot \frac{1}{R}.
In a difference, Q is one quantity and the minus sign acts on all of it. Put the second numerator in brackets before combining, the way Grouping symbols mark a part to be treated as one number, and then remove the brackets by changing the sign of every term inside.
The least common denominator. For number fractions the least common denominator is the least common multiple of the denominators, found from prime factorizations: 12 = 2^2 \cdot 3 and 18 = 2 \cdot 3^2 give 2^2 \cdot 3^2 = 36, each prime taken to the highest power it reaches in either number. Polynomials work the same way, with factors that cannot be factored further in the place of primes. Factor each denominator completely, list every different factor, and take each one to the highest power it has in any single denominator. The product is the least common denominator, the LCD.
A denominator divides the LCD exactly when each of its factors appears in the LCD at least as many times, so the highest power is just enough for all the denominators at once, and anything smaller leaves one of them out. Multiplying the denominators together also gives a common denominator, but a factor they share is then counted twice, and it has to be canceled at the end.
Renaming. Multiply the numerator and the denominator of each fraction by the factors its denominator lacks. That multiplies the fraction by \displaystyle \frac{k}{k} = 1, so its value does not change, which is the same move as renaming number fractions. Keep the LCD in factored form: at the end, that is what shows whether anything cancels.
Combining and simplifying. Add or subtract the numerators over the LCD, multiply out and collect the numerator, then factor it. A factor shared by the numerator and the denominator cancels.
The excluded values. The sum is defined only where every one of the original fractions is. A value that makes any original denominator zero is excluded, and it stays excluded even when its factor cancels at the end. Write the restriction beside the answer.
An expression is not an equation. Multiplying \displaystyle \frac{3}{x + 2} + \frac{5}{x - 1} by (x + 2)(x - 1) gives 3(x - 1) + 5(x + 2), which is a different expression: at x = 0 the sum is \displaystyle -\frac{7}{2} and the new expression is 7. In an equation, multiplying both sides by the same thing keeps it true, and that is how a rational equation is solved. An expression has no second side, so you may not “clear” its denominators. The LCD enters an expression only as \displaystyle \frac{k}{k}.
Part 2: complex fractions
A complex fraction is a fraction whose numerator or denominator, or both, contains a fraction. Simplifying one means writing it as a single rational expression in lowest terms. There are two sound ways.
Method 1: combine, then divide. Write the top as one fraction and the bottom as one fraction, by Part 1. The main fraction bar means division, so then multiply the top by the reciprocal of the bottom, as in dividing rational expressions.
Method 2: multiply top and bottom by the LCD. Find the LCD L of every small fraction in the top and the bottom, and multiply both the top and the bottom of the big fraction by L. Every small denominator divides L, so every small fraction disappears at once. This is allowed because it multiplies the complex fraction by \displaystyle \frac{L}{L} = 1; the forbidden move in Part 1 multiplied by L alone.
Method 1 suits a top and a bottom that are each a single fraction already. Method 2 is quicker when there are many small fractions.
The restrictions. A complex fraction is undefined wherever a small denominator is zero, and also wherever its whole bottom is zero. Both kinds of value stay excluded in the answer.
Worked examples
Examples 1 to 4 go with Part 1. Examples 5 to 7 go with Part 2.
Same denominators
Simplify (a) \displaystyle \frac{x^2}{x + 7} + \frac{7x}{x + 7} \qquad (b) \displaystyle \frac{3x + 1}{x - 4} - \frac{x + 9}{x - 4}
Solution(a) The denominators match, so add the numerators, then factor to look for a shared factor:
The value -7 stays excluded. At x = -7 the original has a zero denominator, while x alone would happily give -7.
(b) Bracket the second numerator, then let the minus sign reach both of its terms, so the +9 becomes -9:
So the difference is 2, for x \ne 4. Check at x = 5: \displaystyle \frac{16}{1} - \frac{14}{1} = 2.
Building the least common denominator
Find the LCD of (a) \displaystyle \frac{1}{6a^2b} and \displaystyle \frac{1}{9ab^3} \qquad (b) \displaystyle \frac{1}{x^2 - x} and \displaystyle \frac{1}{x^2 - 1}, and then add these two \qquad (c) \displaystyle \frac{1}{x^2 + 6x + 9} and \displaystyle \frac{1}{2x + 6}
Solution(a) Factor each: 6a^2b = 2 \cdot 3 \cdot a^2 \cdot b and 9ab^3 = 3^2 \cdot a \cdot b^3. The highest powers are 2, 3^2, a^2 and b^3, so the LCD is 18a^2b^3.
(b) x^2 - x = x(x - 1), and x^2 - 1 = (x - 1)(x + 1) is a difference of squares. The different factors are x, x - 1 and x + 1, none appearing more than once in a single denominator, so the LCD is x(x - 1)(x + 1). The product of the denominators would have carried x - 1 twice.
The first fraction lacks x + 1 and the second lacks x:
so the sum is
The numerator 2x + 1 shares no factor with the denominator, so this is in lowest terms.
(c) x^2 + 6x + 9 = (x + 3)^2 and 2x + 6 = 2(x + 3). The factor x + 3 reaches power 2 in the first, and the 2 appears in the second, so the LCD is 2(x + 3)^2.
Unlike denominators
Simplify (a) \displaystyle \frac{3}{x + 2} + \frac{5}{x - 1} \qquad (b) \displaystyle \frac{x}{x^2 - 9} - \frac{2}{x + 3}
Solution(a) The denominators share no factor, so the LCD is (x + 2)(x - 1). Each fraction gets the factor it lacks:
The numerator has neither x + 2 nor x - 1 as a factor, so this is the answer, for x \ne -2, 1. Check at x = 0: \displaystyle \frac{3}{2} + \frac{5}{-1} = -\frac{7}{2}, and \displaystyle \frac{7}{(2)(-1)} = -\frac{7}{2}.
(b) x^2 - 9 = (x - 3)(x + 3), so that is the LCD, and the second fraction lacks x - 3:
The minus sign acts on all of 2(x - 3) = 2x - 6, which is why the -6 became +6. The numerator 6 - x is -(x - 6), which matches no factor below, so this is in lowest terms, for x \ne 3, -3.
When the result simplifies
Simplify (a) \displaystyle \frac{4}{x^2 - 4} + \frac{1}{x + 2} \qquad (b) \displaystyle \frac{x^2}{x - 6} + \frac{36}{6 - x}
Solution(a) The LCD is (x - 2)(x + 2), and the second fraction lacks x - 2:
for x \ne 2, -2. The shared factor x + 2 only appeared after combining, which is why the LCD was kept factored. And -2 is still excluded: both original denominators are zero there, although \displaystyle \frac{1}{x - 2} is not.
(b) The denominators are opposites: 6 - x = -(x - 6). Multiply the top and bottom of the second fraction by -1, which multiplies it by 1, to get \displaystyle \frac{-36}{x - 6}. Now the denominators match:
A complex fraction, both ways
Simplify, and state the excluded values:
Method 1. Combine the top over its LCD, 4x:
The main bar divides this by x - 4, which is the same as multiplying by \displaystyle \frac{1}{x - 4}. Since 4 - x = -(x - 4),
Method 2. The small denominators are x and 4, with LCD 4x. Multiply the top and the bottom by 4x, distributing over each term of the top: \displaystyle 4x \cdot \frac{1}{x} - 4x \cdot \frac{1}{4} = 4 - x, and the bottom becomes 4x(x - 4). That is the same fraction as in Method 1, and it simplifies to the same \displaystyle -\frac{1}{4x}.
Restrictions. x = 0 makes a small denominator zero, and x = 4 makes the whole bottom zero. So the answer is \displaystyle -\frac{1}{4x} for x \ne 0, 4. The simplified form has a value at x = 4, but the original does not: there it reads \displaystyle \frac{0}{0}.
Check at x = 2: \displaystyle \left(\frac{1}{2} - \frac{1}{4}\right) \div (-2) = -\frac{1}{8}, and \displaystyle -\frac{1}{4 \cdot 2} = -\frac{1}{8}.
Many small fractions
Simplify, and state the excluded values:
There are five terms, three of them fractions, so Method 2 is quicker. The LCD of the small denominators x and x^2 is x^2. Multiply the top and the bottom by x^2, every term, the 1s included:
Now collect the restrictions. The small denominators are zero at x = 0. The whole bottom, \displaystyle 1 - \frac{9}{x^2}, is zero when x^2 = 9, that is at x = 3 and x = -3. So the answer is \displaystyle \frac{x - 1}{x - 3} for x \ne 0, 3, -3.
The simplified form on its own excludes only 3. The other two values leave holes in the graph:
Method 1 gives the same result with more writing: the top is \displaystyle \frac{x^2 + 2x - 3}{x^2}, the bottom is \displaystyle \frac{x^2 - 9}{x^2}, and dividing cancels the two x^2.
Squeezing lemonade together
Jacoba squeezes a jug of lemonade for her stand in t minutes. Hrafn, working alone, takes 6 minutes longer. (a) Write the part of a jug they squeeze together in one minute as a single fraction. (b) Write the time they need together for one jug as a single fraction. (c) Find that time when t = 12.
Solution(a) Jacoba squeezes \displaystyle \frac{1}{t} of a jug a minute and Hrafn \displaystyle \frac{1}{t + 6}. Working together, the unit rates add. The LCD is t(t + 6):
(b) Time for one jug is one jug divided by the rate, which is a complex fraction:
Multiply the top and the bottom by t(t + 6):
That is the answer to (a) turned upside down, as it must be. In this setting t > 0, so no denominator here can be zero.
(c) At t = 12: \displaystyle \frac{12 \cdot 18}{30} = \frac{216}{30} = \frac{36}{5}, which is 7.2 minutes. Check that it is sensible: together they must be quicker than Jacoba’s 12 minutes alone.
Boundaries
No denominator may be zero. Every value that makes an original denominator zero, or for a complex fraction its whole bottom zero, is excluded, and stays excluded after canceling. The simplified answer equals the original only at the allowed values.
Numerators combine only over one shared denominator. The denominators are never added or subtracted themselves; they are made equal first.
The LCD is least only when every denominator is factored completely. A denominator left unfactored, such as x^2 - 1 beside x - 1, still gives a common denominator, but not the least one, and the answer then needs canceling. A factor such as x^2 + 1 that does not factor over the real numbers counts as one factor.
Opposite factors are one factor. x - 6 and 6 - x differ only by a factor of -1, so one of them is renamed and they are counted once.
An expression’s denominators cannot be cleared. Multiplying by the LCD is allowed only in an equation, to both sides, or in a complex fraction, to the top and the bottom together.
Where this goes next
The next step is solving rational equations, where an equation has two sides and clearing denominators becomes the right move. Every answer it gives still has to be tested against the excluded values found the way they are found here.
Running the move backwards splits one fraction into a sum, so \displaystyle \frac{8x + 7}{(x + 2)(x - 1)} goes back to \displaystyle \frac{3}{x + 2} + \frac{5}{x - 1}. That is partial fractions, which calculus uses to integrate rational functions. Complex fractions return in the difference quotient (Exercise 21 is one), and combined rates like Jacoba’s and Hrafn’s are typical rational functions in context.
Exercises
Part 1: a common denominator
Simplify, and state the excluded values.
- \displaystyle \frac{x^2}{x - 8} - \frac{64}{x - 8}
- \displaystyle \frac{4x + 3}{x + 6} - \frac{x - 15}{x + 6}
- \displaystyle \frac{2y - 1}{y^2 + y} + \frac{1 - y}{y^2 + y}
Part 1: the least common denominator
Find the LCD.
- \displaystyle \frac{1}{4m^3n} and \displaystyle \frac{1}{10mn^2}
- \displaystyle \frac{1}{x^2 + 5x} and \displaystyle \frac{1}{x^2 - 25}
- \displaystyle \frac{1}{x^2 - 8x + 16} and \displaystyle \frac{1}{3x - 12}
Part 1: unlike denominators
Simplify, and state the excluded values.
- \displaystyle \frac{2}{x - 5} + \frac{4}{x + 3}
- \displaystyle \frac{6}{x + 4} - \frac{1}{x - 2}
- \displaystyle \frac{7}{2a} + \frac{3}{a^2}
- \displaystyle \frac{x}{x^2 - 16} - \frac{3}{x - 4}
- \displaystyle \frac{x + 1}{x^2 + 3x + 2} + \frac{2}{x + 2}
Part 1: results that simplify
- \displaystyle \frac{10}{x^2 - 25} - \frac{1}{x - 5}
- \displaystyle \frac{2x}{x - 9} + \frac{18}{9 - x}
- \displaystyle \frac{1}{x} + \frac{1}{x + 1} - \frac{1}{x(x + 1)}
- Bronislav spends 24 dollars on a sign and a pitcher for his lemonade stand, and the lemons, sugar and cup for each drink cost \displaystyle \frac{1}{2} dollar. Write his average cost per drink after n drinks, \displaystyle \frac{24}{n} + \frac{1}{2}, as a single fraction, and find it when n = 48.
Part 2: complex fractions
Simplify, and state the excluded values.
- \displaystyle \dfrac{\frac{1}{x} + \frac{1}{3}}{x + 3}
- \displaystyle \dfrac{1 - \frac{4}{x^2}}{1 + \frac{2}{x}}
- \displaystyle \dfrac{\frac{2}{x + 1}}{\frac{4}{x^2 - 1}}
- \displaystyle \dfrac{\frac{3}{x - 2} - 1}{x - 5}
- \displaystyle \dfrac{\frac{a}{b} - \frac{b}{a}}{\frac{1}{a} + \frac{1}{b}}
- \displaystyle \dfrac{\frac{1}{x + h} - \frac{1}{x}}{h}
- Ciprian spends 12 dollars on lemons at p dollars a kilogram one week, and another 12 dollars at q dollars a kilogram the next. His average price is the total spent divided by the total weight, \displaystyle \dfrac{24}{\frac{12}{p} + \frac{12}{q}}. Simplify it, and find it when p = 3 and q = 6.
Reasoning
Exercises 23 and 24 go with Part 1, and Exercise 25 with Part 2.
- Explain why the LCD of \displaystyle \frac{1}{x^2 - 1} and \displaystyle \frac{1}{x + 1} is x^2 - 1 and not the product of the two denominators. What happens if the product is used anyway?
- Fiacre multiplies \displaystyle \frac{2}{x - 5} + \frac{4}{x + 3} by (x - 5)(x + 3) and writes 2(x + 3) + 4(x - 5). Evaluate both at x = 0, and explain why the move is allowed in an equation but not here.
- Explain why multiplying the top and bottom of a complex fraction by the LCD of its small fractions does not change its value, and why x = -2 is excluded from the answer to Exercise 17 even though \displaystyle \frac{x - 2}{x} has a value there.
Answers
- x + 8, x \ne 8. The numerator is x^2 - 64 = (x - 8)(x + 8).
- 3, x \ne -6. (4x + 3) - (x - 15) = 3x + 18 = 3(x + 6).
- \displaystyle \frac{1}{y + 1}, y \ne 0, -1. The numerator is 2y - 1 + 1 - y = y, and y^2 + y = y(y + 1).
- 20m^3n^2. 4m^3n = 2^2 m^3 n and 10mn^2 = 2 \cdot 5 \cdot m n^2; take 2^2, 5, m^3, n^2.
- x(x + 5)(x - 5). The denominators are x(x + 5) and (x - 5)(x + 5).
- 3(x - 4)^2. The denominators are (x - 4)^2 and 3(x - 4).
- \displaystyle \frac{6x - 14}{(x - 5)(x + 3)}, which is also \displaystyle \frac{2(3x - 7)}{(x - 5)(x + 3)}, for x \ne 5, -3. The numerator is 2(x + 3) + 4(x - 5).
- \displaystyle \frac{5x - 16}{(x + 4)(x - 2)}, x \ne -4, 2. The numerator is 6(x - 2) - (x + 4) = 6x - 12 - x - 4.
- \displaystyle \frac{7a + 6}{2a^2}, a \ne 0. The LCD is 2a^2: \displaystyle \frac{7a}{2a^2} + \frac{6}{2a^2}.
- \displaystyle \frac{-2(x + 6)}{(x - 4)(x + 4)}, x \ne 4, -4. The numerator is x - 3(x + 4) = -2x - 12.
- \displaystyle \frac{3}{x + 2}, x \ne -1, -2. Over (x + 1)(x + 2) the numerator is (x + 1) + 2(x + 1) = 3(x + 1), and -1 stays excluded after the canceling.
- \displaystyle -\frac{1}{x + 5}, x \ne 5, -5. The numerator is 10 - (x + 5) = 5 - x = -(x - 5).
- 2, x \ne 9. \displaystyle \frac{18}{9 - x} = \frac{-18}{x - 9}, and 2x - 18 = 2(x - 9).
- \displaystyle \frac{2}{x + 1}, x \ne 0, -1. Over x(x + 1) the numerator is (x + 1) + x - 1 = 2x.
- \displaystyle \frac{n + 48}{2n} dollars, and 1 dollar when n = 48. The LCD is 2n: \displaystyle \frac{48}{2n} + \frac{n}{2n}. At n = 48, \displaystyle \frac{96}{96} = 1.
- \displaystyle \frac{1}{3x}, x \ne 0, -3. The top is \displaystyle \frac{3 + x}{3x}; dividing by x + 3 cancels it.
- \displaystyle \frac{x - 2}{x}, x \ne 0, -2. Multiply by x^2: \displaystyle \frac{x^2 - 4}{x^2 + 2x} = \frac{(x - 2)(x + 2)}{x(x + 2)}. The bottom \displaystyle 1 + \frac{2}{x} is zero at x = -2.
- \displaystyle \frac{x - 1}{2}, x \ne 1, -1. \displaystyle \frac{2}{x + 1} \cdot \frac{(x - 1)(x + 1)}{4}. The bottom \displaystyle \frac{4}{x^2 - 1} is never zero, but its own denominator is zero at \pm 1.
- \displaystyle -\frac{1}{x - 2}, x \ne 2, 5. The top is \displaystyle \frac{3 - (x - 2)}{x - 2} = \frac{5 - x}{x - 2}, and 5 - x = -(x - 5) cancels against the bottom.
- a - b, for a \ne 0, b \ne 0, a \ne -b. Multiply by ab: \displaystyle \frac{a^2 - b^2}{b + a} = \frac{(a - b)(a + b)}{a + b}. The bottom is zero when a = -b.
- \displaystyle -\frac{1}{x(x + h)}, for h \ne 0, x \ne 0, x \ne -h. Multiply by x(x + h): the top becomes x - (x + h) = -h and the bottom hx(x + h).
- \displaystyle \frac{2pq}{p + q} dollars a kilogram, and 4 dollars a kilogram when p = 3, q = 6. Multiply by pq: \displaystyle \frac{24pq}{12q + 12p}. At those prices, \displaystyle \frac{36}{9} = 4; he bought 4 + 2 = 6 kilograms for 24 dollars.
- x^2 - 1 = (x - 1)(x + 1), so x + 1 already divides it: x^2 - 1 is a common denominator, and none can be smaller, since any common denominator must have x^2 - 1 as a factor. The product (x^2 - 1)(x + 1) also works, but it carries x + 1 twice, so the sum comes out with a factor x + 1 above and below that still has to be canceled.
- The sum is \displaystyle -\frac{2}{5} + \frac{4}{3} = \frac{14}{15} at x = 0, while Fiacre’s expression is 6 - 20 = -14. Multiplying by (x - 5)(x + 3) changes the value of an expression. In an equation the same multiplication is done to both sides, so the two sides stay equal; an expression has no other side to balance it.
- Multiplying the top and the bottom by the same non-zero L multiplies the fraction by \displaystyle \frac{L}{L} = 1. At x = -2 the original reads \displaystyle \frac{1 - 1}{1 - 1} = \frac{0}{0}, so it has no value there; the canceled factor x + 2 hides that, and the restriction must be carried over by hand.