Algebraic Fractions

Intuition

A fraction like $\tfrac{6}{9}$ is a ratio of two numbers, and you already know how to shrink it to $\tfrac{2}{3}$ , multiply it, and add it to another. This lesson changes one thing: the top and bottom may now be expressions with a variable, as in $\tfrac{6x^2}{9x}$ or $\tfrac{x + 1}{x - 2}$ . Nothing about the rules changes, because an algebraic fraction is still a ratio, and division still means what it always did. Every move you make here, cancelling, multiplying, flipping to divide, finding a common denominator, is the ordinary fraction rule you learned in arithmetic. You now carry it out with expressions in place of plain numbers. The one new habit is watching the denominator, since a bottom that could turn into zero is a genuine hazard once a variable is present.

Learning goals

Exclude every value that makes the denominator zero
Cancel a shared factor, never a term
Multiply straight across, cancelling first where you can
Divide by multiplying by the reciprocal of the divisor
Add over a common denominator, distributing a leading minus

What an algebraic fraction is

An algebraic fraction, also called a rational expression, is a ratio of two expressions, one written over the other:

\frac{\text{numerator}}{\text{denominator}}, \qquad \text{for example} \qquad \frac{x + 1}{x - 2}.

The bar means exactly what it has always meant, “divide the top by the bottom.” So the fraction stands for the number you get once you choose a value of $x$ , substitute it, and carry out the division. At $x = 5$ the example above is $\tfrac{5 + 1}{5 - 2} = \tfrac{6}{3} = 2$ .

There is one value you are never allowed to choose. Division by zero has no meaning, so the denominator can never equal zero. Any value of the variable that would make the bottom zero is excluded, and the fraction simply is not defined there. In $\tfrac{x + 1}{x - 2}$ the denominator $x - 2$ is zero when $x = 2$ , so $x = 2$ is excluded and we write $x \neq 2$ . Every algebraic fraction carries a restriction like this, and keeping track of it is part of handling the fraction honestly. Finding the excluded values is quick: set the denominator equal to zero and read off the values to avoid. For $\tfrac{5}{x}$ the excluded value is $x = 0$ ; for $\tfrac{x}{2x - 6}$ it is $x = 3$ , since $2x - 6 = 0$ there.

Simplifying by cancelling a common factor

To simplify an algebraic fraction is to rewrite it with a smaller numerator and denominator that name the same value, exactly as $\tfrac{6}{9}$ becomes $\tfrac{2}{3}$ . In arithmetic you did this by dividing top and bottom by a common factor. The rule for expressions is identical, and it rests on a single fact you have used since you first met equivalent fractions. A factor that appears in both the top and the bottom forms a copy of $1$ , and multiplying by $1$ changes nothing.

Why a factor shared by top and bottom can be cancelled#

Suppose the numerator and denominator share a common factor $c$ , so the fraction has the form $\dfrac{a \cdot c}{b \cdot c}$ with $c \neq 0$ and $b \neq 0$ . Because multiplying fractions means multiplying the tops and multiplying the bottoms, that single fraction can be pulled apart into a product of two:

\frac{a \cdot c}{b \cdot c} = \frac{a}{b} \cdot \frac{c}{c}.

Now look at the second factor. Any nonzero quantity divided by itself is $1$ , so $\dfrac{c}{c} = 1$ , and multiplying by $1$ leaves the first factor untouched:

\frac{a}{b} \cdot \frac{c}{c} = \frac{a}{b} \cdot 1 = \frac{a}{b}.

So $\dfrac{a \cdot c}{b \cdot c} = \dfrac{a}{b}$ . Cancelling the common factor $c$ is nothing more than recognizing the hidden $\tfrac{c}{c} = 1$ and dropping it. This is why the two fractions name the same value: they differ only by a factor of $1$ . The one condition is that $c$ must not be zero, which is guaranteed as long as the variable stays away from the excluded values.

The whole method, then, is to factor the numerator and denominator, find the factor they share, and cancel it. The tool for the factoring is the one from the previous lesson: pull out the greatest common factor. Take $\tfrac{6x^2}{9x}$ . Writing $6x^2 = 3x \cdot 2x$ and $9x = 3x \cdot 3$ exposes the shared factor $3x$ , which cancels:

\frac{6x^2}{9x} = \frac{3x \cdot 2x}{3x \cdot 3} = \frac{2x}{3}, \qquad x \neq 0.

Factoring the top and bottom of 6x squared over 9x exposes the shared factor 3x. Because 3x over 3x is 1, that factor cancels and leaves 2x over 3.

The same idea handles a numerator that is a sum, once you factor it. In $\tfrac{x^2 + 3x}{x}$ the top factors as $x(x + 3)$ , so the shared $x$ cancels:

\frac{x^2 + 3x}{x} = \frac{x(x + 3)}{x} = x + 3, \qquad x \neq 0.

And a shared binomial factor cancels the same way a monomial does. Since $3x + 6 = 3(x + 2)$ , the factor $x + 2$ is common to top and bottom of $\tfrac{3x + 6}{x + 2}$ :

\frac{3x + 6}{x + 2} = \frac{3(x + 2)}{x + 2} = 3, \qquad x \neq -2.

Here is the trap that snares more students than any other, so it is worth stating sharply: you may cancel common factors, never individual terms. A factor is something multiplied by the rest; a term is something added to the rest. Cancelling is dividing, and dividing only distributes over a product, not over a sum. In $\tfrac{x + 3}{3}$ the $3$ on the bottom is a factor, but the $3$ on top is a term, added to $x$ , not multiplying it. There is no shared factor, so nothing cancels, and $\tfrac{x + 3}{3}$ is not $x + 1$ . You can check with a value: at $x = 3$ the real fraction is $\tfrac{6}{3} = 2$ , while $x + 1 = 4$ . If you want to split it, the honest split keeps both terms over the $3$ :

\frac{x + 3}{3} = \frac{x}{3} + \frac{3}{3} = \frac{x}{3} + 1.

The same warning applies to $\tfrac{x + 2}{x}$ : the $x$ on the bottom is a factor, but on top it is a term, so the fraction does not become $x + 2$ . Splitting it correctly gives $\tfrac{x}{x} + \tfrac{2}{x} = 1 + \tfrac{2}{x}$ , which is nothing like $x + 2$ .

Worked example 1 Simplify three algebraic fractions

In each, factor the top and bottom, then cancel only a shared factor.

For $\dfrac{6x^2}{9x}$ , the greatest common factor of the two terms is $3x$ , since $6x^2 = 3x \cdot 2x$ and $9x = 3x \cdot 3$ :

\frac{6x^2}{9x} = \frac{3x \cdot 2x}{3x \cdot 3} = \frac{2x}{3}, \qquad x \neq 0.

For $\dfrac{x^2 + 5x}{x}$ , factor the numerator as $x(x + 5)$ and cancel the shared $x$ :

\frac{x^2 + 5x}{x} = \frac{x(x + 5)}{x} = x + 5, \qquad x \neq 0.

For $\dfrac{4x + 8}{x + 2}$ , factor the numerator as $4(x + 2)$ , which shares the factor $x + 2$ with the denominator:

\frac{4x + 8}{x + 2} = \frac{4(x + 2)}{x + 2} = 4, \qquad x \neq -2.

Each answer is the original fraction in lowest terms, valid everywhere except the excluded value.

Check your understanding

Which is the correct simplification of $\dfrac{x + 6}{6}$ ?

x + 1

x

C It is already in simplest form D

\dfrac{x}{6}

Multiplying algebraic fractions

Multiplying algebraic fractions follows the numerical rule with no change: multiply the numerators, multiply the denominators.

\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}.

After multiplying, simplify the result by cancelling any common factor. In practice it is easier to cancel first, before multiplying, because the numbers stay small and the common factors are already sitting in front of you. Take $\tfrac{2x}{3} \cdot \tfrac{9}{x^2}$ . The $9$ on top and $3$ on the bottom share the factor $3$ , and one $x$ on top cancels one of the two on the bottom. So cancelling before multiplying gives

\frac{2x}{3} \cdot \frac{9}{x^2} = \frac{2x \cdot 9}{3 \cdot x^2} = \frac{18x}{3x^2} = \frac{6}{x}, \qquad x \neq 0.

When a numerator or denominator is a sum, factor it first so any shared binomial can cancel.

Worked example 2 Multiply and simplify two products

Multiply tops and bottoms, then cancel every shared factor.

For $\dfrac{x + 2}{5} \cdot \dfrac{10}{x + 2}$ , the binomial $x + 2$ appears on both a top and a bottom, and $\tfrac{10}{5} = 2$ :

\frac{x + 2}{5} \cdot \frac{10}{x + 2} = \frac{10(x + 2)}{5(x + 2)} = \frac{10}{5} = 2, \qquad x \neq -2.

For $\dfrac{3x}{x + 2} \cdot \dfrac{x + 2}{6}$ , cancel the shared $x + 2$ and reduce $\tfrac{3}{6}$ :

\frac{3x}{x + 2} \cdot \frac{x + 2}{6} = \frac{3x(x + 2)}{6(x + 2)} = \frac{3x}{6} = \frac{x}{2}, \qquad x \neq -2.

Cancelling the binomial before multiplying kept both products short.

Dividing algebraic fractions

Dividing looks harder than multiplying, but a single rule turns it back into multiplying: to divide by a fraction, multiply by its reciprocal. The reciprocal of $\tfrac{c}{d}$ is $\tfrac{d}{c}$ , its top and bottom swapped. So

\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}.

This is the same rule you used for numbers, and it holds for the same reason.

Why dividing by a fraction means multiplying by its reciprocal#

Dividing by a quantity means multiplying by whatever undoes it, that is, by its multiplicative inverse: the number you multiply it by to get $1$ . So to divide by $\tfrac{c}{d}$ , we need the fraction that multiplies with $\tfrac{c}{d}$ to give $1$ .

That partner is $\tfrac{d}{c}$ . Multiplying the two, tops together and bottoms together, gives

\frac{c}{d} \cdot \frac{d}{c} = \frac{cd}{dc} = 1,

since the numerator and denominator are the same product. So $\tfrac{d}{c}$ is exactly the inverse of $\tfrac{c}{d}$ , and dividing by $\tfrac{c}{d}$ is multiplying by $\tfrac{d}{c}$ :

\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}.

Nothing here depended on the letters being numbers rather than expressions, so the flip-and-multiply rule carries over to algebraic fractions unchanged.

Once you have flipped the divisor and switched to multiplication, the problem is a multiplication problem. So you finish it the same way: cancel shared factors, then multiply what remains.

Worked example 3 Divide two algebraic fractions

Flip the second fraction, change division to multiplication, then cancel.

For $\dfrac{x}{4} \div \dfrac{x^2}{8}$ , the reciprocal of $\tfrac{x^2}{8}$ is $\tfrac{8}{x^2}$ :

\frac{x}{4} \div \frac{x^2}{8} = \frac{x}{4} \cdot \frac{8}{x^2} = \frac{8x}{4x^2} = \frac{2}{x}, \qquad x \neq 0.

For $\dfrac{3x + 6}{x} \div \dfrac{x + 2}{4x}$ , first factor $3x + 6 = 3(x + 2)$ , then flip and multiply:

\begin{aligned} \frac{3x + 6}{x} \div \frac{x + 2}{4x} &= \frac{3(x + 2)}{x} \cdot \frac{4x}{x + 2} \\ &= \frac{12x(x + 2)}{x(x + 2)} = 12, \qquad x \neq 0, \; x \neq -2. \end{aligned}

The binomial $x + 2$ and the single $x$ both cancel, leaving the plain number $12$ .

Check your understanding

Simplify $\dfrac{2}{x} \div \dfrac{4}{x^2}$ .

\dfrac{x}{2}

\dfrac{8}{x^3}

\dfrac{1}{2x}

2x

Adding and subtracting algebraic fractions

Addition and subtraction are the operations that demand a common denominator, just as they did for numbers. You cannot combine $\tfrac{1}{2}$ and $\tfrac{1}{3}$ until both are written in sixths, and you cannot combine two algebraic fractions until they share a denominator either.

When the denominators are already the same, the work is short: keep the common denominator and combine the numerators. Treat them as the expressions they are and collect like terms.

\frac{a}{d} + \frac{b}{d} = \frac{a + b}{d}, \qquad \frac{a}{d} - \frac{b}{d} = \frac{a - b}{d}.

So $\tfrac{2x}{7} + \tfrac{3x}{7} = \tfrac{5x}{7}$ , adding the like terms $2x$ and $3x$ over the shared $7$ . Subtraction carries the one hazard this lesson keeps returning to. When the numerator being subtracted is a sum, the minus sign applies to every term of it. So wrap that numerator in parentheses and distribute, exactly as you did when subtracting whole expressions:

\frac{4x + 1}{x} - \frac{x - 2}{x} = \frac{(4x + 1) - (x - 2)}{x} = \frac{4x + 1 - x + 2}{x} = \frac{3x + 3}{x}.

The $-2$ became $+2$ ; missing that flip is the classic slip.

When the denominators differ, first rewrite each fraction as an equivalent one over a common denominator, then combine. The smallest denominator that every fraction divides into is the least common denominator (LCD). Building each equivalent fraction uses the same equivalent-fraction rule as always: multiply top and bottom by whatever the denominator is missing.

Worked example 4 Combine fractions that already share a denominator

Keep the common denominator and combine the numerators, watching the sign on any subtraction.

Adding over the shared $x + 1$ , the numerators are like terms:

\frac{5x}{x + 1} + \frac{2x}{x + 1} = \frac{5x + 2x}{x + 1} = \frac{7x}{x + 1}, \qquad x \neq -1.

Subtracting, the minus sign reaches both terms of the second numerator, so distribute it:

\begin{aligned} \frac{6x + 5}{x + 2} - \frac{2x - 3}{x + 2} &= \frac{(6x + 5) - (2x - 3)}{x + 2} \\ &= \frac{6x + 5 - 2x + 3}{x + 2} = \frac{4x + 8}{x + 2}, \qquad x \neq -2. \end{aligned}

The $-3$ turned into $+3$ , giving the constant $5 + 3 = 8$ ; treating the subtraction as reaching only the $2x$ would have produced the wrong $4x + 2$ .

Worked example 5 Combine fractions with different denominators

Rewrite each fraction over the least common denominator, then add the numerators.

For $\dfrac{x}{4} + \dfrac{x}{6}$ , the LCD of $4$ and $6$ is $12$ . Multiply the first fraction by $\tfrac{3}{3}$ and the second by $\tfrac{2}{2}$ :

\frac{x}{4} + \frac{x}{6} = \frac{3x}{12} + \frac{2x}{12} = \frac{5x}{12}.

For $\dfrac{3}{x} + \dfrac{1}{2}$ , the LCD of $x$ and $2$ is $2x$ . Multiply the first by $\tfrac{2}{2}$ and the second by $\tfrac{x}{x}$ :

\frac{3}{x} + \frac{1}{2} = \frac{6}{2x} + \frac{x}{2x} = \frac{6 + x}{2x}, \qquad x \neq 0.

The numerator $6 + x$ has unlike terms, so it cannot be combined further, and the sum stays as one fraction over the common denominator.

Check your understanding

Simplify $\dfrac{5x}{x + 2} - \dfrac{2x - 3}{x + 2}$ .

\dfrac{3x + 3}{x + 2}

\dfrac{3x - 3}{x + 2}

\dfrac{7x - 3}{x + 2}

3x + 3

Common mistakes

Cancelling a term instead of a factor. You may cancel only a factor shared by the whole top and the whole bottom. So $\tfrac{x + 3}{3}$ is not $x + 1$ , and $\tfrac{x + 2}{x}$ is not $x + 2$ . Factor first; if there is no shared factor, nothing cancels.
Forgetting the denominator cannot be zero. Every algebraic fraction excludes the values that make its denominator zero. In $\tfrac{x + 1}{x - 2}$ the value $x = 2$ is off limits, and a simplified form like $\tfrac{3x + 6}{x + 2} = 3$ still carries the restriction $x \neq -2$ .
Dropping a sign when subtracting a sum. A minus sign in front of a fraction reaches every term of its numerator. So $\tfrac{4x + 1}{x} - \tfrac{x - 2}{x} = \tfrac{3x + 3}{x}$ , not $\tfrac{3x - 1}{x}$ ; the $-2$ becomes $+2$ .
Adding without a common denominator. $\tfrac{3}{x} + \tfrac{1}{2}$ is not $\tfrac{4}{x + 2}$ or $\tfrac{4}{2x}$ . Rewrite each fraction over the LCD $2x$ first, giving $\tfrac{6 + x}{2x}$ .
Adding denominators when multiplying. To multiply, multiply straight across: $\tfrac{a}{b} \cdot \tfrac{c}{d} = \tfrac{ac}{bd}$ . A common denominator is needed for adding, not for multiplying.
Flipping the wrong fraction when dividing. Reciprocate the divisor (the fraction after the division sign), never the first one: $\tfrac{a}{b} \div \tfrac{c}{d} = \tfrac{a}{b} \cdot \tfrac{d}{c}$ .

Key takeaways

An algebraic fraction is a ratio of two expressions. Its denominator can never be zero, so every value of the variable that makes the bottom zero is excluded (in $\tfrac{x + 1}{x - 2}$ , $x \neq 2$ ).
Simplify by factoring the top and bottom and cancelling a shared factor, which is valid because $\tfrac{c}{c} = 1$ : $\tfrac{6x^2}{9x} = \tfrac{2x}{3}$ for $x \neq 0$ , and $\tfrac{3x + 6}{x + 2} = 3$ for $x \neq -2$ .
Cancel factors, never terms. $\tfrac{x + 3}{3}$ is not $x + 1$ , because the $3$ on top is a term, not a factor.
Multiply straight across, then simplify (cancel first when you can): $\tfrac{a}{b} \cdot \tfrac{c}{d} = \tfrac{ac}{bd}$ .
Divide by multiplying by the reciprocal of the divisor, because $\tfrac{c}{d} \cdot \tfrac{d}{c} = 1$ : $\tfrac{a}{b} \div \tfrac{c}{d} = \tfrac{a}{b} \cdot \tfrac{d}{c}$ .
Add or subtract over a common denominator, combining like terms in the numerator; when subtracting a sum, distribute the minus to every term: $\tfrac{6x + 5}{x + 2} - \tfrac{2x - 3}{x + 2} = \tfrac{4x + 8}{x + 2}$ .

Algebraic Fractions

Learning goals

What an algebraic fraction is

Simplifying by cancelling a common factor

Multiplying algebraic fractions

Dividing algebraic fractions

Adding and subtracting algebraic fractions

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)