Simplifying rational expressions

A rational expression is a quotient of polynomials, i.e. an algebraic fraction: (x + 1)/(x² − 4), 3/a + a/b and so on. Working with them is like working with ordinary fractions: cancelling, finding a common denominator, multiplying and dividing are done exactly as with numerical fractions. But there are two extra requirements: the domain (the denominator must not be zero) and factorising (only factors can be cancelled). This topic is needed constantly in equations, inequalities and functions, and later in limits and integrals.

§1. Rational expressions and their domain

A polynomial makes sense for every x. A fractional expression makes sense only where the denominator is not zero, because division by zero is undefined.

How to find it: set the denominator equal to zero, solve, and exclude the roots.

• For (x + 1)/(x² − 4): x² − 4 ≠ 0 → x ≠ ±2.
• For 5/(x² + 1) there is no restriction: x² + 1 > 0 always.
• For 1/x + 2/(x − 3): x ≠ 0 and x ≠ 3.

Important: the restrictions remain after cancelling. (x² − 1)/(x² + x) = (x − 1)/x, but the original expression is also undefined at x = −1, so the answer is (x − 1)/x, with x ≠ 0 and x ≠ −1. Two expressions are equal only on their common domain.

§2. The fundamental property of a fraction and cancelling

Cancelling is this property used in reverse. Procedure:

1. Factorise the numerator and denominator (take out a common factor, use special products, factorise a quadratic).
2. Find the common factor and divide by it.

Why? (x + 3)/3 = x/3 + 1 — the whole numerator is divided by 3, not just part of it.
A useful observation: (b − a) = −(a − b), so (a − b)/(b − a) = −1.
A quadratic factorises as ax² + bx + c = a(x − x₁)(x − x₂), where x₁, x₂ are its roots.

Worked example

Simplify (a² − 9)/(a² − 6a + 9).

1. Numerator: a² − 9 = (a − 3)(a + 3).
2. Denominator: a² − 6a + 9 = (a − 3)².
3. Cancel (a − 3): (a + 3)/(a − 3), with a ≠ 3.

§3. Adding and subtracting

Fractions with the same denominator are added by adding numerators. Fractions with different denominators are first brought to the lowest common denominator:

1. Factorise each denominator.
2. The common denominator is the product of all factors that occur, each to its highest power.
3. For each fraction find the extra factor and multiply its numerator by it.
4. Add the numerators and cancel the result if possible.

When subtracting, the numerator of the second fraction must be put in brackets — otherwise a sign error appears.

Worked example

Simplify 1/(x − 1) − 1/(x + 1).

1. Common denominator: (x − 1)(x + 1) = x² − 1.
2. Extra factors: x + 1 for the first fraction, x − 1 for the second.
3. ((x + 1) − (x − 1))/(x² − 1) = 2/(x² − 1).

Worked example

Simplify x/(x − 2) − 8/(x² − 4).

1. x² − 4 = (x − 2)(x + 2) is the common denominator.
2. The extra factor for the first fraction is x + 2: (x(x + 2) − 8)/((x − 2)(x + 2)).
3. Numerator: x² + 2x − 8 = (x + 4)(x − 2).
4. Cancel (x − 2): (x + 4)/(x + 2), with x ≠ ±2.

§4. Multiplying, dividing and compound fractions

It is convenient to cancel before multiplying: factorise everything first, then cross out common factors. When dividing, the second fraction is turned upside down (replaced by its reciprocal).
To simplify a compound fraction, first write its numerator and denominator each as a single fraction, then divide. For example, (1 + 1/x)/(1 − 1/x²) = ((x + 1)/x) · (x²/((x − 1)(x + 1))) = x/(x − 1).

Worked example

Simplify (a/b − b/a) : (a + b)/(ab).

1. Inside the bracket: a/b − b/a = (a² − b²)/(ab).
2. Replace division with multiplication: (a² − b²)/(ab) · ab/(a + b).
3. (a − b)(a + b)/(a + b) = a − b.

§5. Values under a condition

Sometimes the value of an expression must be found without finding x and y. The idea is to rewrite the expression so that the given condition appears in it:

• If x − y = 5: 6x + 5 − 6y = 6(x − y) + 5 = 35.
• If a/b = 3: a = 3b, (2a + b)/(a − b) = 7b/(2b) = 7/2.
• In a homogeneous fraction (all terms of the numerator and denominator have the same degree) divide top and bottom by b and write everything in terms of a/b.

§6. Rational equations and substitution

How to solve a rational equation:

1. Write down the domain (denominators ≠ 0).
2. Multiply both sides by the common denominator to get a polynomial equation.
3. Solve that equation.
4. Compare the roots with the domain and reject extraneous roots.

Why does an extraneous root appear? We multiplied by an expression that can be zero: at x = 2, x² − 4 = 0, so the multiplied equation holds while the original one is meaningless.
For symmetric expressions, introducing a new variable makes the work easier.

Worked example

Solve x/(x + 2) + 4/(x² − 4) = 1/(x − 2).

1. Domain: x ≠ ±2.
2. Multiply by x² − 4: x(x − 2) + 4 = x + 2.
3. x² − 3x + 2 = 0 → x = 1 or x = 2.
4. x = 2 is outside the domain. Answer: x = 1.

Worked example

Find the sum of the real roots of x²/3 + 48/x² = 10(x/3 − 4/x).

1. Let t = x/3 − 4/x; then t² = x²/9 − 8/3 + 16/x².
2. Left side: x²/3 + 48/x² = 3(x²/9 + 16/x²) = 3(t² + 8/3) = 3t² + 8.
3. 3t² + 8 = 10t → 3t² − 10t + 8 = 0 → t = 2 or t = 4/3.
4. t = 2: x² − 6x − 12 = 0 (D > 0), sum of roots 6.
5. t = 4/3: x² − 4x − 12 = 0 → x = 6; −2, sum 4.
6. Total: 6 + 4 = 10 (x = 0 is not among the roots).

Check yourself

1. Find the domain of (x + 1)/(x² − 4).

2. Simplify (x² − 1)/(x² + x).

3. Simplify 1/(x − 1) − 2/(x² − 1).

4. Find (2a + b)/(a − b) if a/b = 3.

5. Solve (x² − 4)/(x − 2) = 5.

6. Solve x/(x − 3) = 3/(x − 3).

7. Solve 3/(x − 1) + 2/(x + 1) = 4/(x² − 1).

8. Simplify (1 + 1/x)/(1 − 1/x²).

Common mistakes

• Cancelling terms: (x + 3)/3 ≠ x + 1.
• Forgetting the domain after cancelling.
• Including an extraneous root (one that makes a denominator zero) in the answer to a rational equation.
• Not changing every sign in the numerator of the second fraction when subtracting.
• Using the product of the denominators instead of the lowest common denominator and making the work needlessly complicated.
• Turning the first fraction upside down when dividing (it is the divisor that is inverted).