Learning Objectives
List all excluded inputs before simplifying rational expressions.
Add, multiply and divide algebraic fractions using factors.
Distinguish equivalent values on a common domain from identical domains.
Check candidate solutions of rational equations against restrictions.
Prerequisites: Section 1B Lesson 01 expansion, factorisation and identities; exact fraction arithmetic. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
Rational expressions are algebraic fractions: their numerators and denominators can depend on a variable. Their arithmetic follows the familiar rules for numerical fractions, with one extra task: keeping track of the values that make a denominator or divisor zero.
This lesson begins with the domain, then develops addition, multiplication and division. We finish with rational equations, where cancelling or clearing a denominator must not introduce an answer that was forbidden in the original problem.
1. The Domain of an Algebraic Fraction
Before changing an algebraic fraction, identify where it has a value. This list of allowed inputs is its domain. Writing the exclusions first gives you a reference that remains valid even when factors disappear during simplification.
1.1 Rational expressions and denominator zeros
A rational expression is a quotient of polynomials, such as (x + 1)/(x² − 4). A polynomial is a finite sum of constant multiples of nonnegative integer powers of x. Its denominator cannot be zero, so this example requires x ≠ 2 and x ≠ −2.
1.2 Simplification: equal values on the original domain
Factor the original denominator to find its zeros before cancelling. The simplified formula may be defined at more inputs than the original. Equality of values wherever the original exists does not enlarge the original domain.
The point of cancellation is to compute an existing value more simply. It does not assign a new value to a quotient that originally involved division by zero. Keep the excluded inputs beside the simplified formula.
Worked Example: A Cancelled Restriction Remains
Question: Simplify (x² − x − 6)/(x² − 4), stating all restrictions.
Method: Factor numerator and denominator, record their denominator zeros, and cancel only on that original domain.
Factor: (x − 3)(x + 2)/[(x − 2)(x + 2)]. The original denominator excludes x = 2 and x = −2.
For the allowed inputs, cancel x + 2 to get (x − 3)/(x − 2). Keep x ≠ ±2. At x = 0, both expressions are 3/2; at x = −2, only the new formula has a value, so that input is still excluded from the original.
2. Adding and Subtracting Algebraic Fractions
Just as 1/2 and 1/3 must be expressed using equal-sized parts before adding, algebraic fractions need a shared denominator. Factoring shows which pieces are already shared and which multipliers are still needed.
2.1 Equivalent fractions: forming a common denominator
Addition needs equivalent fractions with a common denominator, not a product of numerators. Factor the denominators first so you can use their shared factors efficiently. Put brackets around an entire numerator when subtracting a fraction.
Multiplying numerator and denominator by the same nonzero expression leaves a fraction unchanged on its allowed domain. Use this to give both fractions the same denominator before adding or subtracting the numerators.
For nonzero denominators A and B, p/A + q/B = (pB + qA)/(AB). If the denominators share factors, a smaller common denominator may be available. Whatever denominator you use, retain every original restriction.
2.2 Subtraction: distributing the minus sign over the numerator
The numerator of the fraction being subtracted is one grouped expression. For example, subtracting (x − 1) means adding −x + 1, not −x − 1. Keep its brackets until that minus sign has been applied to every term.
Worked Example: Subtract the Whole Numerator
Question: Simplify 2/(x − 1) − 1/(x + 1).
Method: Use the product of the two linear denominators and subtract the entire second numerator in brackets.
The restrictions are x ≠ 1 and x ≠ −1. Use (x − 1)(x + 1): the numerator is 2(x + 1) − (x − 1) = x + 3.
Thus the answer is (x + 3)/(x² − 1), with both restrictions. At x = 2 the original is 2 − 1/3 = 5/3, matching 5/3 from the result.
3. Multiplication and Division: Checking Different Conditions
Products and quotients can both be simplified by factoring, but division asks an additional question: does the fraction you are dividing by have value zero? This section separates that extra restriction from the original denominator restrictions.
3.1 Multiplying fractions: cancel shared nonzero factors
To multiply fractions, factor their numerators and denominators, then cancel shared nonzero factors. There is no need to expand large products first. A cancellation is justified only after the original denominators have been checked.
3.2 Dividing fractions: the divisor must exist and be nonzero
To divide by a fraction, multiply by its reciprocal, but the divisor must itself be both defined and nonzero. This adds a restriction at any zero of the divisor numerator that lies in its domain. Flipping a fraction does not erase those conditions.
A fraction with numerator zero and nonzero denominator has value zero. It is perfectly valid as a factor in a multiplication, but invalid as a divisor. This is why a division can exclude an input that a corresponding multiplication would allow.
Worked Example: The Divisor Must Not Vanish
Question: Simplify [(x² − 1)/(x − 2)] ÷ [(x + 1)/(x − 3)].
Method: Check both fraction denominators and the divisor’s numerator before multiplying by the reciprocal.
The two original denominators exclude 2 and 3. The divisor is zero at x = −1, so exclude that too.
Multiply by the reciprocal: [(x − 1)(x + 1)/(x − 2)] × [(x − 3)/(x + 1)]. Cancel to obtain (x − 1)(x − 3)/(x − 2), for x ≠ −1, 2, 3.
At x = 0, the original is (1/2) ÷ (−1/3) = −3/2, agreeing with the result. The cancellation does not make x = −1 a valid input.
4. Rational Equations: Checking the Candidate Answers
A rational expression asks for a simplified formula; a rational equation asks which inputs make an equality true. The domain still comes first, because clearing denominators is only a reversible step where the common denominator is nonzero.
4.1 Clearing denominators on the allowed domain
On the allowed domain, multiplying an equation by a nonzero common denominator gives an equivalent equation. Solve the resulting equation and then reject any candidate outside the original domain. If you ignore this final step, a cleared equation can appear to have a solution where the original is undefined.
4.2 No solutions or every allowed input: interpreting the result
A rational equation may also hold at every point of its domain. “All allowed real x” is different from “all real x” when there are exclusions.
After solving, compare each candidate with the exclusions you recorded. If the only candidate is forbidden, there is no solution. If the equation simplifies to an identity, describe its solutions using the original allowed domain rather than ignoring the excluded points.
Worked Example: A Candidate That Cannot Be Used
Question: Solve (x² − 1)/(x − 1) = 2 for real x.
Method: Simplify on the allowed domain, solve the resulting linear equation, and compare its candidate with the original exclusion.
The domain is x ≠ 1. On that domain the left side is x + 1, so any solution would satisfy x + 1 = 2, giving x = 1.
This candidate is excluded. The original equation has no solution; substituting x = 1 gives a zero denominator, not a valid equality.
Common Mistakes
Cancel factors of the whole numerator and denominator, never selected terms in a sum.
Subtract a bracketed numerator so every sign changes correctly.
When dividing by a fraction, exclude its zeros as well as its undefined inputs.
Carry restrictions into the final answer even if the simplified formula looks harmless there.
5. Practice
Begin by writing the domain for each fraction or equation. The first four questions practise the arithmetic operations; the final three ask you to interpret the resulting solution set. Keep restrictions visible throughout your working.
1. Simplify (x² − 5x + 6)/(x − 2) and state its domain.
2. Simplify 1/x + 1/(x + 2), with restrictions.
3. Simplify [x/(x − 1)] × [(x² − 1)/x²], with restrictions.
4. Simplify [x/(x − 1)] ÷ [(x + 2)/(x − 1)], with restrictions.
5. Solve 2/(x − 1) = 1/(x + 1).
6. Solve (x² − 4)/(x − 2) = 4.
7. For how many real x is (x² − 9)/(x − 3) = x + 3 true? A: none. B: one. C: every real x except 3. D: every real x. Explain.
Worked Solutions
1. Factor and retain the denominator exclusion
Factor the numerator as (x − 2)(x − 3). The result is x − 3 for x ≠ 2. At x = 0 both expressions equal −3; x = 2 remains forbidden.
2. Add using a common denominator
The numerator over x(x + 2) is (x + 2) + x = 2x + 2. The result is 2(x + 1)/[x(x + 2)], for x ≠ 0, −2. At x = 1 both sides are 4/3.
3. Cancel only after listing original restrictions
Factor x² − 1 = (x − 1)(x + 1). Cancellation gives (x + 1)/x, with x ≠ 0, 1 from the original denominators. At x = 2 both forms equal 3/2.
4. Exclude a zero divisor before taking its reciprocal
The original fractions require x ≠ 1, and the divisor must not be zero, so x ≠ −2. Multiplying by the reciprocal gives x/(x + 2), with x ≠ 1, −2. The value x = 0 is allowed and gives 0 in both forms.
5. Clear denominators and check the candidate
Exclude x = ±1. Multiplication by (x − 1)(x + 1) gives 2(x + 1) = x − 1, hence x = −3. Check: 2/(−4) = 1/(−2) = −1/2.
6. Reject an excluded solution
Exclude x = 2. Simplification gives x + 2 = 4, whose only candidate is the excluded value 2. There is no solution.
7. Distinguish an identity on a domain from an unrestricted identity
C. Factor the numerator as (x − 3)(x + 3). Cancellation proves equality for every x ≠ 3, giving infinitely many solutions. At 3 the original quotient is undefined, so D is false.
6. Summary
- Begin with the original domain and preserve it.
- Use common denominators for sums and factor cancellation for products.
- A divisor must be defined and nonzero before you take its reciprocal.
- Check every candidate solution in the original rational equation.