Algebraic fractions are expressions in which polynomials appear in both the numerator and the denominator. They represent a natural extension of numerical fractions: just as a numerical fraction expresses the ratio of two numbers, an algebraic fraction expresses the ratio of two algebraic expressions.
However, unlike numerical fractions, algebraic fractions require additional care: the denominator may depend on one or more variables and can therefore equal zero for certain values. For this reason, it is not enough to know how to calculate; one must first determine for which values the expression is defined.
Contents
- What is an algebraic fraction
- Domain restrictions and domain
- Equivalent algebraic fractions
- Simplifying algebraic fractions
- Finding a common denominator
- Operations with algebraic fractions
- Expressions involving algebraic fractions
- Equations with algebraic fractions
- Common mistakes to avoid
What is an algebraic fraction
An algebraic fraction is an expression of the form
\[ \frac{A}{B} \]
where \(A\) and \(B\) are algebraic expressions with \(B\neq 0\). In the most common cases, \(A\) and \(B\) are polynomials. The expression \(A\) is called the numerator and \(B\) is called the denominator.
For example,
\[ \frac{x+1}{x-2}, \qquad \frac{x^2-1}{x^2+3x+2}, \qquad \frac{2a-b}{a^2-b^2} \]
are algebraic fractions.
The denominator is the central feature of the theory. Indeed, a fraction — whether numerical or algebraic — is undefined whenever the denominator equals zero. For this reason, before transforming or simplifying an algebraic fraction, one must identify all values of the variable for which the denominator vanishes.
Domain restrictions and domain
An algebraic fraction
\[ \frac{A(x)}{B(x)} \]
is defined for all values of the variable for which the denominator is nonzero:
\[ B(x)\neq 0. \]
This condition is called the domain restriction. The set of all values satisfying this condition is the domain of the algebraic fraction.
Example
Consider the fraction
\[ \frac{x+3}{x-5}. \]
The denominator is \(x-5\). We require it to be nonzero:
\[ x-5\neq 0, \]
which gives
\[ x\neq 5. \]
Therefore the fraction is defined for all real values of \(x\) except \(5\). The domain is
\[ \mathbb{R}\setminus\{5\}. \]
Example with a factorable denominator
Consider
\[ \frac{x+1}{x^2-4}. \]
The denominator equals zero when
\[ x^2-4=0. \]
Since
\[ x^2-4=(x-2)(x+2), \]
we require
\[ (x-2)(x+2)\neq 0. \]
A product is nonzero if and only if each factor is nonzero. Therefore
\[ x\neq 2 \quad \text{and} \quad x\neq -2. \]
The domain is
\[ \mathbb{R}\setminus\{-2,2\}. \]
Equivalent algebraic fractions
Two algebraic fractions are equivalent if they take the same value for every element of their common domain.
The fundamental property is analogous to that of numerical fractions: multiplying both numerator and denominator by the same nonzero expression yields an equivalent fraction.
If \(C\neq 0\), then
\[ \frac{A}{B}=\frac{A\cdot C}{B\cdot C}, \qquad B\neq 0,\ C\neq 0. \]
This property underlies both simplification and the process of finding a common denominator. The condition \(C\neq 0\) is not a mere formality: multiplying or dividing by an expression that may vanish risks altering the domain of the fraction.
Simplifying algebraic fractions
Simplifying an algebraic fraction means dividing both numerator and denominator by a common nonzero factor. To do this correctly, both numerator and denominator must first be factored completely.
Terms connected by addition or subtraction cannot be cancelled. Only common factors may be cancelled.
Example
Simplify
\[ \frac{x^2-1}{x^2+2x+1}. \]
We begin by factoring numerator and denominator:
\[ x^2-1=(x-1)(x+1), \]
\[ x^2+2x+1=(x+1)^2. \]
Therefore
\[ \frac{x^2-1}{x^2+2x+1} = \frac{(x-1)(x+1)}{(x+1)^2}. \]
The factor \(x+1\) is common to both numerator and denominator. We may cancel it, provided we note that \(x+1\neq 0\), i.e. \(x\neq -1\):
\[ \frac{(x-1)(x+1)}{(x+1)^2} = \frac{x-1}{x+1}. \]
Hence
\[ \frac{x^2-1}{x^2+2x+1} = \frac{x-1}{x+1}, \qquad x\neq -1. \]
It is important to observe that the simplified fraction equals the original only on the domain of the original fraction. Simplification does not permit one to discard the domain restrictions.
Finding a common denominator
To add or subtract algebraic fractions, they must be rewritten with a common denominator. The most convenient choice is usually the least common multiple of the denominators, computed after factoring each one completely.
The procedure is as follows:
- factor each denominator completely;
- determine the least common denominator;
- rewrite each fraction as an equivalent fraction with that denominator;
- add or subtract the numerators.
Example
Rewrite the following fractions with a common denominator:
\[ \frac{1}{x-1} \quad \text{and} \quad \frac{2}{x+1}. \]
The denominators are already fully factored. The least common denominator is
\[ (x-1)(x+1). \]
Therefore
\[ \frac{1}{x-1} = \frac{x+1}{(x-1)(x+1)} \]
and
\[ \frac{2}{x+1} = \frac{2(x-1)}{(x-1)(x+1)}. \]
The domain restrictions are
\[ x\neq 1, \qquad x\neq -1. \]
Operations with algebraic fractions
Addition and subtraction
To add or subtract two algebraic fractions with the same denominator, add or subtract the numerators and keep the denominator:
\[ \frac{A}{B}+\frac{C}{B} = \frac{A+C}{B}, \qquad B\neq 0. \]
Similarly,
\[ \frac{A}{B}-\frac{C}{B} = \frac{A-C}{B}, \qquad B\neq 0. \]
Example
Compute
\[ \frac{1}{x-1}+\frac{2}{x+1}. \]
The common denominator is \((x-1)(x+1)\). Therefore
\[ \frac{1}{x-1}+\frac{2}{x+1} = \frac{x+1}{(x-1)(x+1)} + \frac{2(x-1)}{(x-1)(x+1)}. \]
We can now add the numerators:
\[ \frac{x+1+2(x-1)}{(x-1)(x+1)}. \]
Expanding the numerator:
\[ x+1+2x-2=3x-1. \]
Therefore
\[ \frac{1}{x-1}+\frac{2}{x+1} = \frac{3x-1}{(x-1)(x+1)}. \]
The domain restrictions are
\[ x\neq 1,\qquad x\neq -1. \]
Multiplication
The product of two algebraic fractions is obtained by multiplying the numerators together and the denominators together:
\[ \frac{A}{B}\cdot \frac{C}{D} = \frac{AC}{BD}, \qquad B\neq 0,\ D\neq 0. \]
Before carrying out the multiplication it is often convenient to factor and cancel any common factors.
Example
Compute
\[ \frac{x^2-1}{x^2}\cdot \frac{x}{x+1}. \]
Factoring:
\[ x^2-1=(x-1)(x+1). \]
Therefore
\[ \frac{x^2-1}{x^2}\cdot \frac{x}{x+1} = \frac{(x-1)(x+1)}{x^2}\cdot \frac{x}{x+1}. \]
We can cancel the factor \(x+1\) and one factor of \(x\), noting the restrictions \(x\neq 0\) and \(x\neq -1\):
\[ \frac{(x-1)(x+1)}{x^2}\cdot \frac{x}{x+1} = \frac{x-1}{x}. \]
Hence
\[ \frac{x^2-1}{x^2}\cdot \frac{x}{x+1} = \frac{x-1}{x}, \qquad x\neq 0,\ x\neq -1. \]
Division
Dividing by an algebraic fraction means multiplying by its reciprocal, provided the divisor is nonzero:
\[ \frac{A}{B}:\frac{C}{D} = \frac{A}{B}\cdot \frac{D}{C} = \frac{AD}{BC}. \]
The conditions are
\[ B\neq 0,\qquad D\neq 0,\qquad C\neq 0. \]
The condition \(C\neq 0\) is necessary because the fraction \(\frac{C}{D}\), being the divisor, cannot equal zero.
Expressions involving algebraic fractions
When evaluating expressions that contain algebraic fractions, it is advisable to proceed in an orderly fashion. The domain restrictions are established first; then the operations are carried out in accordance with the standard order of precedence, respecting parentheses.
Example
Simplify
\[ \left(\frac{x}{x-1}-\frac{1}{x+1}\right)\cdot \frac{x^2-1}{x}. \]
We first determine the domain restrictions:
\[ x-1\neq 0,\qquad x+1\neq 0,\qquad x\neq 0, \]
that is,
\[ x\neq 1,\qquad x\neq -1,\qquad x\neq 0. \]
We now work on the expression in parentheses:
\[ \frac{x}{x-1}-\frac{1}{x+1}. \]
The common denominator is \((x-1)(x+1)\), so
\[ \frac{x}{x-1} = \frac{x(x+1)}{(x-1)(x+1)} \]
and
\[ \frac{1}{x+1} = \frac{x-1}{(x-1)(x+1)}. \]
Therefore
\[ \frac{x}{x-1}-\frac{1}{x+1} = \frac{x(x+1)-(x-1)}{(x-1)(x+1)}. \]
Expanding the numerator:
\[ x(x+1)-(x-1)=x^2+x-x+1=x^2+1. \]
The full expression then becomes
\[ \frac{x^2+1}{(x-1)(x+1)}\cdot \frac{x^2-1}{x}. \]
Since
\[ x^2-1=(x-1)(x+1), \]
we have
\[ \frac{x^2+1}{(x-1)(x+1)}\cdot \frac{(x-1)(x+1)}{x}. \]
Cancelling the common factors,
\[ \frac{x^2+1}{x}. \]
Therefore
\[ \left(\frac{x}{x-1}-\frac{1}{x+1}\right)\cdot \frac{x^2-1}{x} = \frac{x^2+1}{x}, \qquad x\neq -1,\ 0,\ 1. \]
Equations with algebraic fractions
Equations containing algebraic fractions are often called rational equations. Solving them requires particular care, since not every solution obtained algebraically is necessarily valid.
The correct procedure is:
- determine the domain restrictions;
- solve the equation within those restrictions;
- discard any value that makes one of the original denominators equal to zero.
Example
Solve
\[ \frac{x+1}{x-2}=3. \]
The domain restriction is
\[ x-2\neq 0, \]
that is,
\[ x\neq 2. \]
We multiply both sides by \(x-2\), which is nonzero on the domain of the equation:
\[ x+1=3(x-2). \]
Expanding:
\[ x+1=3x-6. \]
Collecting the terms in \(x\) on one side and the constants on the other:
\[ 1+6=3x-x. \]
Therefore
\[ 7=2x, \]
giving
\[ x=\frac{7}{2}. \]
Since \(\frac{7}{2}\neq 2\), this solution is valid:
\[ S=\left\{\frac{7}{2}\right\}. \]
Example with an extraneous solution
Solve
\[ \frac{x}{x-1}=\frac{1}{x-1}. \]
The domain restriction is
\[ x-1\neq 0, \]
that is,
\[ x\neq 1. \]
Since both fractions share the same denominator, which is nonzero on the domain, we may equate the numerators:
\[ x=1. \]
However, \(x=1\) does not satisfy the domain restriction, since it makes the denominator equal to zero. This value must therefore be discarded.
The equation has no solution:
\[ S=\varnothing. \]
Common mistakes to avoid
The first mistake is cancelling terms rather than factors. For example,
\[ \frac{x+2}{x} \]
cannot be simplified by cancelling \(x\), because \(x\) is not a factor of the entire numerator: it appears only as a term in the sum \(x+2\).
The second mistake is forgetting the domain restrictions after simplification. For example,
\[ \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1, \]
but this equality holds only for
\[ x\neq 1. \]
Indeed, the original fraction is undefined at \(x=1\), whereas the expression \(x+1\) would be defined there. As expressions with a specified domain, the two are not identical unless the restriction \(x\neq 1\) is retained.
The third mistake is multiplying both sides of an equation by an expression that may be zero without first establishing the domain. In a rational equation, every transformation must be justified within the domain restrictions.
Algebraic fractions are not simply fractions with letters. They are rational expressions whose meaning depends essentially on the denominator. For this reason, every calculation must be accompanied by a check of the domain restrictions.
Simplifying, adding, multiplying, or dividing algebraic fractions means applying the same properties as for numerical fractions, but with greater attention to the domain. The fundamental rule is always the same: expressions may be transformed only in ways that are consistent with the values for which they are defined.
A thorough understanding of algebraic fractions is indispensable for tackling rational equations, rational inequalities, rational functions, and many subsequent topics in algebra and mathematical analysis.