Understand · explore · practise
Simplify, multiply, divide, add and subtract algebraic fractions. Keep track of excluded values, explore cancellation and practise with fully worked solutions.
Before you startFactorising quadratics, difference of two squares and numerical fractions
01 / The essentials
Algebraic fractions obey the same rules as numerical fractions. The extra task is to record which inputs make a denominator zero. Work over the real numbers unless a question says otherwise.
Factor → record restrictions → simplify → check
Keep the original restrictions even when a factor disappears from the final formula.
Before multiplying or dividing, look for common factors. Before adding or subtracting, find a common denominator. In a division, also exclude inputs that make the whole divisor zero.
This lesson extends the Pure 1 fraction rules to more involved expressions and applications. Each method and solution stays readable without playing a video.
02 / Factor and cancel
Cancellation divides the entire numerator and denominator by the same nonzero factor. Factor first so you can see what multiplies the whole expression.
(x² − 9)/(x² − x − 6)
= (x − 3)(x + 3)/[(x − 3)(x + 2)]
= (x + 3)/(x + 2), x ≠ 3, −2
The cancelled factor x − 3 is nonzero only when x ≠ 3. The remaining denominator also needs x ≠ −2.
x is not a factor of the whole numerator. Instead, split the numerator: (x + 6)/x = 1 + 6/x for x ≠ 0. At x = 2 the answer is 4, not 7 or 6.
(2x² − 8)/(x² + x − 6)
Original denominator: (x + 3)(x − 2), so x ≠ −3, 2.
= 2(x − 2)(x + 2)
/[(x + 3)(x − 2)]
Only the shared factor x − 2 cancels.
= 2(x + 2)/(x + 3)
x ≠ −3, 2
Write the simplified form together with its original restrictions.
03 / Keep the domain
The formula G(x) = (x + 3)/(x + 2) gives a value at x = 3. But F(x) = (x² − 9)/(x² − x − 6) gives 0/0 there, which is undefined. Thus F and G have equal values wherever F is defined, but their natural domains differ.
Writing “F(x) = G(x), for x ≠ −2, 3” is correct. Writing “F(3) = 6/5” is not. Simplifying cannot change the meaning of the original expression.
Use the control to visit allowed and excluded inputs. The readout explains the two different reasons an input can fail.
F(x) = (x² − 9)/(x² − x − 6)
G(x) = (x + 3)/(x + 2)
x = 0: F(x) = 3/2 and G(x) = 3/2. Both are defined and equal.
Both original factors in the denominator are nonzero.
Try x = −2, then x = 3. At −2 both expressions are undefined. At 3 only G exists: G(3) = 6/5. The original function still has a missing input.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
04 / Multiply
Multiply numerators together and denominators together. You may cancel a factor from one numerator against a factor in either denominator, provided you retain the restrictions of the original product.
(a/b) × (c/d) = ac/(bd)
Requires b ≠ 0 and d ≠ 0. Numerators are allowed to be zero.
(x² − 16)/(3x)
× 6x²/(x + 4)
Original restrictions: x ≠ 0, −4.
= (x − 4)(x + 4) · 6x²
/[3x(x + 4)]
Cancel x + 4; reduce 6/3 and x²/x.
= 2x(x − 4), x ≠ 0, −4
No visible denominator remains, but both restrictions stay. At x = 4 the product is zero and is allowed.
05 / Divide
Invert the second fraction and multiply. Record conditions before taking the reciprocal: each original denominator is nonzero, and the numerator of the divisor is nonzero.
(a/b) ÷ (c/d) = ad/(bc)
Requires b ≠ 0, d ≠ 0 and c ≠ 0.
An input may be invalid because the divisor is undefined, or because it equals zero. Those are different failures; both must be excluded.
(x² − 1)/(x + 3)
÷ (x − 1)/(x + 2)
x ≠ −3, −2 from the denominators. Also x ≠ 1 because the divisor would be zero.
= (x − 1)(x + 1)/(x + 3)
× (x + 2)/(x − 1)
Invert the whole divisor. Now cancel x − 1.
= (x + 1)(x + 2)/(x + 3)
x ≠ −3, −2, 1
At x = −2 the simplified formula gives zero, but the original divisor is undefined.
06 / Add fractions
Rewrite both fractions over a common denominator before adding their numerators. Denominators do not add. If one denominator already contains the other, use the larger factor combination rather than duplicating factors unnecessarily.
2/(x − 1) + 3/(x + 2)
x ≠ 1, −2. Common denominator: (x − 1)(x + 2).
= [2(x + 2) + 3(x − 1)]
/[(x − 1)(x + 2)]
Multiply top and bottom of each fraction by its missing factor.
= (5x + 1)/[(x − 1)(x + 2)]
Expand only the numerator; the factored denominator makes the restrictions easy to see.
1/x − 2/[x(x + 1)]
x ≠ 0, −1. Common denominator: x(x + 1).
= [(x + 1) − 2]/[x(x + 1)]
Only the first fraction needs an extra factor.
= (x − 1)/[x(x + 1)]
No common factor remains. Keep x ≠ 0, −1.
07 / Subtract fractions
Put brackets around the second numerator after forming a common denominator. Expand it first, then change every sign when you subtract.
(x + 3)/(x − 2)
− (x − 1)/(x + 2)
x ≠ 2, −2.
= [(x + 3)(x + 2)
− (x − 1)(x − 2)]/(x² − 4)
The entire second product is subtracted.
Numerator = x² + 5x + 6
− (x² − 3x + 2)
= 8x + 4
The last two terms become +3x and −2.
Result: (8x + 4)/(x² − 4)
x ≠ −2, 2
A numerical check such as x = 0 gives −1 from both forms. It helps catch a sign error but does not replace the algebra.
08 / Use the result
Equations: (x² − 9)/(x − 3) = 6 requires x ≠ 3. Cancelling gives x + 3 = 6 and candidate x = 3. Reject it: the original equation has no solution.
Logarithms: solve log₂[(x² − 4)/(x − 2)] = 3. The fraction requires x ≠ 2; after simplification its value is x + 2, which must be positive. Thus x > −2 and x ≠ 2. Now x + 2 = 2³ = 8 gives x = 6, which satisfies both conditions.
Differentiation: if f(x) = (x³ − 8)/(x − 2), factor the difference of cubes to get f(x) = x² + 2x + 4 for x ≠ 2. Hence f′(x) = 2x + 2 for x ≠ 2. The original f is not defined at 2, so it has no derivative there.
Around each allowed input, the rational expression and polynomial agree on a small interval. Their rates of change therefore agree there. Adding a new value at the missing input would define a different, extended function.
09 / Your turn
Try each question independently. Your answer is incomplete without excluded inputs. Hints and full solutions are separate so you can choose how much support to reveal.
Simplify (x² − 25)/(x² + 2x − 15).
Factor the denominator into two linear factors.
The denominator is (x + 5)(x − 3), so x ≠ −5, 3. The numerator is (x − 5)(x + 5). Cancel x + 5 to obtain (x − 5)/(x − 3), still with x ≠ −5, 3.
Simplify (x² − 6x + 9)/(x² − 9).
The numerator is a perfect square.
Write (x − 3)²/[(x − 3)(x + 3)], with x ≠ 3, −3. Cancel one copy of x − 3: (x − 3)/(x + 3), keeping both exclusions.
Simplify [(x² − 4)/(5x)] × [10x²/(x + 2)].
Factor x² − 4 before multiplying.
Original restrictions are x ≠ 0, −2. The product is (x − 2)(x + 2)·10x²/[5x(x + 2)] = 2x(x − 2), for x ≠ 0, −2. A polynomial answer does not remove the original exclusions.
Simplify [(x² − 16)/(x + 1)] ÷ [(x − 4)/(x − 2)].
Exclude the zero of the divisor as well as both original denominator zeros.
x ≠ −1, 2, 4. Invert and multiply: (x − 4)(x + 4)(x − 2)/[(x + 1)(x − 4)] = (x + 4)(x − 2)/(x + 1), with all three exclusions.
Simplify 1/(2x − 1) + 2/(x + 3).
Use (2x − 1)(x + 3) as the common denominator.
x ≠ 1/2, −3. The numerator becomes (x + 3) + 2(2x − 1) = 5x + 1. Answer: (5x + 1)/[(2x − 1)(x + 3)].
Simplify 3/[x(x − 1)] − 1/(x − 1).
The second numerator must be multiplied by x.
x ≠ 0, 1. Over x(x − 1), the numerator is 3 − x. Answer: (3 − x)/[x(x − 1)], for x ≠ 0, 1.
Simplify (1/x + 1/3)/(1/x − 1/3).
x cannot be zero. Also the whole bottom expression must not be zero.
x ≠ 0, and 1/x − 1/3 ≠ 0 gives x ≠ 3. The top is (3 + x)/(3x), and the bottom is (3 − x)/(3x). Their quotient is (3 + x)/(3 − x), with x ≠ 0, 3.
Solve (x² − 4)/(x − 2) = x² − 4.
First exclude x = 2; then simplify the left-hand side.
For x ≠ 2, x + 2 = x² − 4. Thus x² − x − 6 = (x − 3)(x + 2) = 0. Candidates x = 3 and x = −2 are both allowed and satisfy the original equation. Both are solutions.
Find the real domain of ln[(x² − 9)/(x − 3)].
The original denominator must be nonzero and the argument of ln must be positive.
x ≠ 3. On that domain the argument equals x + 3, so require x + 3 > 0. The domain is x > −3, with x ≠ 3; equivalently (−3, 3) ∪ (3, ∞).
For f(x) = (x³ − 27)/(x − 3), find f′(x) and explain whether f′(3) exists.
Use x³ − 27 = (x − 3)(x² + 3x + 9).
For x ≠ 3, f(x) = x² + 3x + 9, so f′(x) = 2x + 3. Since the original f is undefined at 3, f′(3) does not exist. Substituting 3 into the derivative formula would ignore its domain.
10 / Recap
Section 1 of 10 · The essentials