Adding and Subtracting Rational Expressions

Intuition

A fraction is a count of pieces. The expression $\frac{3}{7}$ means three pieces, each one seventh of a whole. The sum $\frac{3}{7} + \frac{2}{7} = \frac{5}{7}$ holds because three sevenths and two sevenths are counts of the same size of piece, so the counts add. That is also why $\frac{1}{2} + \frac{1}{3}$ cannot be done as it stands: halves and thirds are different sizes, and adding $1 + 1$ would be adding counts of two different things. Nothing in that reasoning cares whether the denominator is the number $7$ or the polynomial $x - 7$ . Every method in this lesson is the fraction method you already own, carried across to polynomials. The one genuinely new feature is that a polynomial denominator can equal zero, so you have to keep track of which values of $x$ are shut out.

Learning goals

Add only over a shared denominator
Build the LCD from each factor at its highest power
Bracket the whole numerator being subtracted, then distribute
Keep restrictions from the original denominators, cancelled or not
Turn a complex fraction into a division

Why a common denominator is needed at all

A common denominator is not a ritual step. It is the only way to make the two counts comparable. Once the pieces are the same size, addition is just counting them together:

\frac{3}{7} + \frac{2}{7} = \frac{5}{7}.

Written out with the size of the piece made explicit, the reason is visible. Three sevenths is $3 \cdot \frac{1}{7}$ and two sevenths is $2 \cdot \frac{1}{7}$ , so the sum is $(3 + 2) \cdot \frac{1}{7}$ . That step is the distributive property, run backwards, and it is the whole justification.

Fractions with the same denominator add by adding numerators#

Let $A$ , $B$ , and $D$ be polynomials, and fix any value of $x$ with $D \neq 0$ , so that all the fractions below are defined. Since dividing by $D$ is the same as multiplying by the number $\frac{1}{D}$ , we can write $\frac{A}{D} = A \cdot \frac{1}{D}$ and $\frac{B}{D} = B \cdot \frac{1}{D}$ . Then

\frac{A}{D} + \frac{B}{D} = A \cdot \frac{1}{D} + B \cdot \frac{1}{D} = (A + B) \cdot \frac{1}{D} = \frac{A + B}{D}.

The middle equality is the distributive property with the common factor $\frac{1}{D}$ pulled out of both terms. Notice what the argument never used: it never asked what $A$ , $B$ , or $D$ actually are. It needed only that $D \neq 0$ , so that $\frac{1}{D}$ exists. That is exactly why the rule for numbers and the rule for polynomials are the same rule, and why nothing about this chapter’s algebra is new.

Replacing $B$ with $-B$ gives the subtraction version, $\frac{A}{D} - \frac{B}{D} = \frac{A - B}{D}$ , with the same restriction $D \neq 0$ .

So the rule needs one denominator. That makes the entire job of adding unlike fractions a single task: rewrite them so that they share a denominator. The tool for that is the other fact you already have, that multiplying the top and the bottom by the same nonzero quantity leaves a fraction’s value alone, because

\frac{C \cdot K}{D \cdot K} = \frac{C}{D} \cdot \frac{K}{K} = \frac{C}{D} \cdot 1 = \frac{C}{D}, \qquad D \neq 0,\; K \neq 0.

Adding when the denominators already match

When the denominators are identical, combine the numerators over that denominator and then simplify the result the way you learned in the first lesson of this chapter. Here that comes down to two moves: factor the numerator and the denominator, and cancel any factor they share.

\frac{A}{D} + \frac{B}{D} = \frac{A + B}{D}, \qquad \frac{A}{D} - \frac{B}{D} = \frac{A - B}{D}.

Worked example 1 Combine $\frac{x^2}{x+5} + \frac{3x-10}{x+5}$

The denominators already match, so put the two numerators over the single denominator:

\frac{x^2}{x+5} + \frac{3x-10}{x+5} = \frac{x^2 + 3x - 10}{x+5}.

Now factor the numerator and look for a factor shared with the denominator. Two numbers multiplying to $-10$ and adding to $3$ are $5$ and $-2$ , so $x^2 + 3x - 10 = (x+5)(x-2)$ :

\frac{(x+5)(x-2)}{x+5} = x - 2.

The answer is $x - 2$ , valid for $x \neq -5$ . The polynomial $x - 2$ is perfectly happy at $x = -5$ , but the expression you started with is not, since $x + 5$ sits in its denominator. Cancelling a factor never widens the domain, so the restriction travels with the answer.

The minus sign owns the whole numerator

Almost every real error in this lesson is a sign error, and it comes from one place. In $\frac{A}{D} - \frac{B}{D} = \frac{A - B}{D}$ , the minus sign applies to all of $B$ , not just to its first term. When $B$ is a single term like $3$ , you can hardly go wrong. The moment $B$ is a polynomial like $3x + 4$ , the minus has to reach both terms, and the $+4$ becomes $-4$ .

The habit that prevents this: write brackets around the numerator you are subtracting before you do anything else with it. Then distribute the minus sign as a separate, deliberate step.

Worked example 2 Combine $\frac{5x-2}{x^2-9} - \frac{3x+4}{x^2-9}$

The denominators match, so subtract the numerators. Bracket the numerator being subtracted:

\frac{5x-2}{x^2-9} - \frac{3x+4}{x^2-9} = \frac{(5x-2) - (3x+4)}{x^2-9}.

Now distribute the minus sign across both terms of $3x + 4$ :

(5x-2) - (3x+4) = 5x - 2 - 3x - 4 = 2x - 6.

Factor the top and the bottom, and cancel:

\frac{2x-6}{x^2-9} = \frac{2(x-3)}{(x-3)(x+3)} = \frac{2}{x+3}.

The restrictions come from the original denominator $x^2 - 9 = (x-3)(x+3)$ , so $x \neq 3$ and $x \neq -3$ . The factor $x - 3$ cancelled, but $x = 3$ is still excluded.

Compare that with the wrong turn. A student who drops the brackets writes $5x - 2 - 3x + 4 = 2x + 2$ and reports $\frac{2x+2}{(x-3)(x+3)}$ , which will not simplify at all. The lost sign on the $+4$ turned $2x - 6$ into $2x + 2$ and destroyed the common factor. A subtraction whose answer refuses to simplify is worth a second look at the sign.

Check your understanding

Simplify $\dfrac{4x+3}{x+2} - \dfrac{2x-1}{x+2}$ .

\dfrac{2x+2}{x+2}

2

\dfrac{6x+2}{x+2}

\dfrac{2}{x+2}

Building the least common denominator

When the denominators differ, you must rebuild both fractions over a shared denominator. Any common denominator will do, but one of them is cheapest, and it is worth knowing which.

Start with numbers, where you already know the answer. To add $\frac{1}{12} + \frac{1}{18}$ , look at what the denominators are made of:

12 = 2^2 \cdot 3, \qquad 18 = 2 \cdot 3^2.

A common denominator has to be a multiple of both. To be a multiple of $12$ it needs two factors of $2$ ; to be a multiple of $18$ it needs two factors of $3$ . Take exactly the highest power of each prime that appears in either one, and you get the least common multiple:

2^2 \cdot 3^2 = 36.

Then $\frac{1}{12} = \frac{3}{36}$ and $\frac{1}{18} = \frac{2}{36}$ , so the sum is $\frac{5}{36}$ and no cancelling is needed at the end.

Polynomials factor into irreducible pieces in essentially one way, just as integers factor into primes, so the same recipe transfers word for word with “prime” replaced by “irreducible factor”. To build the least common denominator (LCD):

Factor every denominator completely.
List the distinct factors that appear anywhere.
Raise each one to the highest power it reaches in any single denominator.
Multiply those together.

Plain numbers left sitting in front of a denominator are handled by the rule you already know: take their ordinary least common multiple. The denominators $2x$ and $3x$ share the factor $x$ , and $\operatorname{lcm}(2,3) = 6$ , so their LCD is $6x$ .

That last clause is also where the word “least” runs out of room. A nonzero number never changes where a denominator vanishes, so if $L$ is a common denominator then so are $2L$ and $3L$ , and they have exactly the same degree. The recipe therefore pins the LCD down only up to a numeric constant. Convention settles the remaining choice in the obvious way: take the version carrying no surplus number, which is what the least common multiple of the coefficients gives you.

Take the denominators $x^2 - 4$ and $x^2 - 4x + 4$ . Factored, they are $(x-2)(x+2)$ and $(x-2)^2$ . The distinct factors are $x - 2$ and $x + 2$ . The highest power of $x - 2$ in a single denominator is $2$ , and the highest power of $x + 2$ is $1$ , so

\text{LCD} = (x-2)^2(x+2).

Each denominator is a stack of factors. The LCD takes every distinct factor as many times as it appears in the denominator that uses it most, and no more.

The recipe is not folklore. It is forced, and here is the argument.

The recipe gives a common denominator, and the smallest one#

Call the product produced by the recipe $L$ .

First, $L$ really is a common denominator. Take any one of the denominators, call it $D$ . Every factor of $D$ appears in $L$ raised to at least the power it has in $D$ . That is because $L$ uses the highest power that factor reaches in any denominator, and $D$ is one of them. So $D$ divides $L$ , which means $L = D \cdot Q$ for some polynomial $Q$ . Multiplying that fraction’s numerator and denominator by $Q$ rewrites it over $L$ , and the same works for every fraction in the sum.

Second, nothing smaller works. Suppose $C$ is any common denominator, so every denominator divides $C$ . Pick one distinct factor $p$ , and let $k$ be the highest power of $p$ occurring in a single denominator, reached in some denominator $D_j$ . Then $p^k$ divides $D_j$ , and $D_j$ divides $C$ , so $p^k$ divides $C$ . Run that argument once for each distinct factor. The distinct factors share no piece with one another, so $C$ must be divisible by the product of all of those highest powers, and that product is exactly $L$ . Hence $L$ divides $C$ , and no common denominator can have smaller degree than $L$ . That is what makes $L$ least, and it is as far as “least” can be pushed: $2L$ and $3L$ are common denominators of the same degree. So the LCD is fixed only up to a numeric constant, and by convention you take the one carrying no surplus number.

Both halves lean on the same fact: a polynomial factors into irreducible pieces in essentially one way, the same pieces every time, each pinned down up to a constant. That is what makes “the highest power of each factor” a well-defined thing to ask for. This is the argument that makes $\operatorname{lcm}(12, 18) = 36$ , transplanted from primes to irreducible polynomials.

Check your understanding

What is the least common denominator of $\dfrac{5}{x^2-4x}$ and $\dfrac{1}{x^2-16}$ ?

x(x+4)

(x-4)(x+4)

x(x-4)(x+4)

x(x-4)^2(x+4)

Why the product is usually the worse choice

Multiplying the denominators together always produces a common denominator, and when they share no factor at all, that product is the LCD. But when they do share a factor, the product carries a copy of it that you will have to cancel back out at the end. Watch the same problem run both ways.

Combine $\dfrac{2}{x^2-4} + \dfrac{3}{x^2-4x+4}$ , whose denominators factor as $(x-2)(x+2)$ and $(x-2)^2$ .

Using the LCD $(x-2)^2(x+2)$ . The first fraction is missing one $x - 2$ and the second is missing one $x + 2$ :

\frac{2}{(x-2)(x+2)} \cdot \frac{x-2}{x-2} + \frac{3}{(x-2)^2} \cdot \frac{x+2}{x+2} = \frac{2(x-2) + 3(x+2)}{(x-2)^2(x+2)}.

The numerator is $2x - 4 + 3x + 6 = 5x + 2$ , so the answer is

\frac{5x+2}{(x-2)^2(x+2)}, \qquad x \neq 2,\; x \neq -2.

Using the product $(x-2)^3(x+2)$ . Now the first fraction needs $(x-2)^2$ and the second needs $(x-2)(x+2)$ :

\frac{2(x-2)^2 + 3(x-2)(x+2)}{(x-2)^3(x+2)} = \frac{2x^2 - 8x + 8 + 3x^2 - 12}{(x-2)^3(x+2)} = \frac{5x^2 - 8x - 4}{(x-2)^3(x+2)}.

That numerator is not in lowest terms, so you are forced to factor a quadratic that you created yourself, $5x^2 - 8x - 4 = (5x+2)(x-2)$ , and cancel:

\frac{(5x+2)(x-2)}{(x-2)^3(x+2)} = \frac{5x+2}{(x-2)^2(x+2)}.

Same answer, more work, and a real risk of stopping early and calling $\frac{5x^2-8x-4}{(x-2)^3(x+2)}$ the final answer. The LCD route never built the surplus $x - 2$ in the first place, so there was nothing to cancel.

The full method, with unlike denominators

Putting the pieces together gives one procedure that always works.

Factor every denominator completely.
Read off the restrictions from those factors, before anything cancels.
Build the LCD: each distinct factor, to the highest power it reaches in any single denominator.
Rewrite each fraction over the LCD by multiplying its numerator and its denominator by the factors it is missing.
Combine the numerators over the single denominator, bracketing every numerator that is being subtracted.
Expand, collect, factor, cancel. The restrictions from step 2 stay with the answer.

Worked example 3 Combine $\frac{x+4}{x^2+3x} - \frac{2}{x+3}$

Factor the denominators: $x^2 + 3x = x(x+3)$ , and the second is already $x + 3$ . So the restrictions are $x \neq 0$ and $x \neq -3$ .

The distinct factors are $x$ and $x + 3$ , each to the first power, so the LCD is $x(x+3)$ . The first fraction already sits over it. The second is missing a factor of $x$ :

\frac{2}{x+3} \cdot \frac{x}{x} = \frac{2x}{x(x+3)}.

Combine, bracketing the numerator that is being subtracted:

\frac{(x+4) - 2x}{x(x+3)} = \frac{4 - x}{x(x+3)}.

The numerator $4 - x$ is zero only at $x = 4$ , while the denominator’s factors vanish at $x = 0$ and $x = -3$ . So the numerator shares no factor with the denominator, and the expression is already in lowest terms:

\frac{4-x}{x(x+3)}, \qquad x \neq 0,\; x \neq -3.

Worked example 4 Combine $\frac{x}{x-2} - \frac{4}{x^2-2x}$

Factor: $x^2 - 2x = x(x-2)$ . The denominators are $x - 2$ and $x(x-2)$ , so the restrictions are $x \neq 2$ and $x \neq 0$ . The LCD is $x(x-2)$ , since that already contains both.

The second fraction sits over the LCD. The first is missing a factor of $x$ :

\frac{x}{x-2} \cdot \frac{x}{x} = \frac{x^2}{x(x-2)}.

Combine over the single denominator:

\frac{x^2 - 4}{x(x-2)} = \frac{(x-2)(x+2)}{x(x-2)} = \frac{x+2}{x}.

The answer is $\frac{x+2}{x}$ , with $x \neq 0$ and $x \neq 2$ . That second restriction is the one people lose. The final form $\frac{x+2}{x}$ shows no trouble at $x = 2$ , but the expression you were handed has $x - 2$ in a denominator. No amount of later algebra can make that expression exist at $x = 2$ .

The restrictions come from the original denominators

The domain is settled before you do any algebra at all. Factor the denominators exactly as they are given, and exclude every value that kills any factor. Two consequences catch people out.

A factor that cancels still restricts. In Worked Example 4 the factor $x - 2$ cancelled, but the starting expression was undefined at $x = 2$ , so the answer is undefined there too. The simplified form agrees with the original at every value it is allowed to have, and says nothing about the value it was never allowed to have.

Building up never adds a new restriction. Every factor you multiply a fraction by, on its way to the LCD, is already a factor of some denominator in the problem. So the values that factor kills were excluded from the very beginning. That has a useful consequence: the LCD is zero at exactly the excluded values, and nowhere else. If some denominator is zero at $x = a$ , then since that denominator divides the LCD, the LCD is zero at $x = a$ too. Going the other way, if the LCD is zero at $x = a$ , then one of its irreducible factors is zero there. That factor came from some denominator, which is therefore also zero at $x = a$ . So once you have the factored LCD in front of you, you can read the full list of restrictions straight off it.

Check your understanding

The expression $\dfrac{3}{x-4} - \dfrac{12}{x^2-4x}$ simplifies to $\dfrac{3}{x}$ . Which values of $x$ must be excluded?

x \neq 0

and

x \neq 4

x \neq 0

only C

x \neq 4

only D

x \neq 0

and

x \neq -4

Complex fractions are division in disguise

A complex fraction has a fraction inside its numerator, inside its denominator, or both. There is nothing new to learn here, because the main fraction bar means division, and dividing rational expressions is already in your hands: multiply by the reciprocal.

Combine the top into a single fraction.
Combine the bottom into a single fraction.
Multiply the top by the reciprocal of the bottom.
Collect restrictions from every denominator that appeared, including the main one, since the bottom of the big fraction cannot be zero either.

Worked example 5 Simplify $\dfrac{\frac{1}{x} + \frac{1}{2}}{\frac{1}{x} - \frac{1}{2}}$

Combine the top over the LCD $2x$ , then the bottom over the same LCD:

\frac{1}{x} + \frac{1}{2} = \frac{2}{2x} + \frac{x}{2x} = \frac{2 + x}{2x}, \qquad \frac{1}{x} - \frac{1}{2} = \frac{2}{2x} - \frac{x}{2x} = \frac{2 - x}{2x}.

Now the complex fraction is one fraction divided by another, so multiply by the reciprocal:

\frac{2+x}{2x} \div \frac{2-x}{2x} = \frac{2+x}{2x} \cdot \frac{2x}{2-x} = \frac{x+2}{2-x}.

For the restrictions, hunt through every denominator that appeared. The $\frac{1}{x}$ terms force $x \neq 0$ . The bottom of the big fraction cannot be zero, and $\frac{2-x}{2x} = 0$ exactly when $x = 2$ , so $x \neq 2$ as well:

\frac{x+2}{2-x}, \qquad x \neq 0,\; x \neq 2.

Notice that the final form still shows $x \neq 2$ in its denominator, but has lost all trace of $x \neq 0$ . That is the restriction you must carry by hand.

Common mistakes

Adding the denominators too. There is no rule $\frac{a}{b} + \frac{c}{d} = \frac{a+c}{b+d}$ . Test it on numbers you know: $\frac{1}{2} + \frac{1}{2}$ is $1$ , not $\frac{2}{4}$ . Denominators name the size of the piece, and adding two sizes together names nothing.
Letting the minus sign reach only the first term. In $\frac{5x-2}{x^2-9} - \frac{3x+4}{x^2-9}$ , the subtraction applies to the $3x$ and the $+4$ . The numerator is $5x - 2 - 3x - 4 = 2x - 6$ , not $2x + 2$ . Bracket the numerator you are subtracting before you touch it.
Multiplying only the denominator when building up. Rewriting $\frac{3}{x-1}$ over $(x-1)(x+4)$ means multiplying top and bottom by $x+4$ , giving $\frac{3x+12}{(x-1)(x+4)}$ . Changing the bottom alone changes the value of the expression.
Cancelling a term instead of a factor. After you combine, $\frac{x+3}{x+5}$ is finished. You cannot cancel the $x$ ‘s, because they are terms, not factors. Only whole factors of the entire top and bottom cancel.
Reading the restrictions off the final answer. They come from the original denominators. In Worked Example 4 the answer $\frac{x+2}{x}$ looks fine at $x = 2$ , but $x = 2$ was excluded before the algebra began and stays excluded.
Reaching for the product of the denominators out of habit. It is a legal common denominator, but when the denominators share a factor it forces you to expand a bigger numerator and then factor it back down. Factor first, build the LCD, and skip the round trip.

Key takeaways

Fractions add only when the pieces are the same size: $\frac{A}{D} \pm \frac{B}{D} = \frac{A \pm B}{D}$ wherever $D \neq 0$ , and the reason is the distributive property, nothing more.
To change a denominator, multiply the numerator and the denominator by the same nonzero factor. That leaves the value alone.
LCD recipe: factor every denominator, then take each distinct factor to the highest power it reaches in any single denominator, and multiply. Numeric coefficients take their ordinary least common multiple. This is the least common multiple, with irreducible polynomials in place of primes.
The product of the denominators is always a common denominator, but when they share a factor it forces a cancellation at the end that the LCD avoids.
Subtraction: bracket the whole numerator being subtracted, then distribute the minus sign as its own step. Most wrong answers in this lesson are a lost sign on a trailing term.
Restrictions come from the original denominators. A factor that cancels still restricts, and the factored LCD is zero at exactly the excluded values.
A complex fraction is a division: combine the top, combine the bottom, multiply by the reciprocal, and collect restrictions from every denominator in sight.

Adding and Subtracting Rational Expressions

Learning goals

Why a common denominator is needed at all

Adding when the denominators already match

The minus sign owns the whole numerator

Building the least common denominator

Why the product is usually the worse choice

The full method, with unlike denominators

The restrictions come from the original denominators

Complex fractions are division in disguise

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)