Adding and Subtracting Algebraic Fractions
Adding and subtracting algebraic fractions follows the same steps as ordinary fractions: find a common denominator, rewrite each fraction over it, then combine the numerators. In NSW Year 10 Advanced maths the denominators contain pronumerals and often need factorising first, as in \(\dfrac{3}{x^2-1}-\dfrac{1}{x+1}\) (NESA outcome MA5‑ALG‑P‑02).
You will start with numerical and single-term denominators, move to binomial denominators such as \((x+4)(x-3)\), spot that \(5-x=-(x-5)\), and factorise quadratic denominators to find the lowest common denominator. Harder questions simplify a fraction inside a fraction, find unknown constants by matching coefficients, and split each term of a long sum so that most of it cancels. Worded questions add the travel times for the legs of a trip. It is part of Year 10 Advanced Algebra Review.
Every question with a fully worked solution.
- Adding and Subtracting Algebraic Fractions - Video - Adding Algebraic Fractions Watch
Theory
To add or subtract algebraic fractions, write every fraction over the lowest common denominator, then combine the numerators. This Year 10 Advanced maths page covers numerical and pronumeral denominators, binomial and quadratic denominators that must be factorised first, reversed brackets such as \(5-x\), and cancelling the answer.
An algebraic fraction has a pronumeral in its numerator, its denominator or both, such as \(\dfrac{x+3}{4}\) or \(\dfrac{2}{x-3}\). The rules are the rules of ordinary fractions.
You can only add or subtract fractions with the same denominator. The lowest common denominator (LCD) is the simplest expression that every denominator divides into. For \(6a\) and \(9a^2\) it is \(18a^2\): the LCM of the numbers and the highest power of each pronumeral.
When the denominators are binomials such as \(x+4\) and \(x-3\), with no common factor, the LCD is their product \((x+4)(x-3)\). When a denominator is a quadratic, factorise it first. A factor that appears in two denominators goes into the LCD only once.
Leave the denominator in factorised form, and factorise the final numerator too. A common factor of the numerator and denominator can then be cancelled. The answer is only valid for values of the pronumeral that do not make a denominator zero.
Same denominator, then combine numerators:
Factorisations you will need for denominators:
| Type | Pattern | Example |
|---|---|---|
| Difference of two squares | \(a^2-b^2 = (a-b)(a+b)\) | \(x^2-16 = (x-4)(x+4)\) |
| Monic quadratic | \(x^2+bx+c = (x+m)(x+n)\), \(mn = c\), \(m+n = b\) | \(x^2-2x-15 = (x-5)(x+3)\) |
| Reversed bracket | \(b-a = -(a-b)\) | \(\dfrac{4}{5-x} = -\dfrac{4}{x-5}\) |
| Perfect square | \((a\pm b)^2 = a^2\pm 2ab+b^2\) | \((x-3)^2 = x^2-6x+9\) |
How to add or subtract algebraic fractions
- Factorise every denominator fully. Check for a reversed bracket such as \(2-x\).
- Cancel inside any single fraction first if you can. It often makes the denominators match.
- Find the LCD: each different factor once, at its highest power.
- Rewrite each fraction over the LCD: multiply its numerator by the factors it is missing.
- Combine the numerators in brackets, then expand carefully. A minus sign multiplies every term of the bracket after it.
- Simplify: collect like terms, factorise the numerator and cancel any common factor. Leave the denominator factorised unless asked otherwise.
Other question types
- Three fractions: the LCD holds every factor of all three denominators, such as \(x(x+1)(x+2)\). Build each numerator separately, then combine.
- Find the constants: if \(\dfrac{p}{x+1}+\dfrac{q}{x-2}\) must equal a given fraction, combine the left side, expand its numerator and match the coefficients of \(x\) and the constant terms.
- Show that: start from one side and work step by step to the exact form asked for. Do not start from the answer.
- Models: time \(=\) distance \(\div\) speed gives one fraction per leg of a trip. Add them, simplify, then substitute (Example 4).
(i) \(\dfrac{x+3}{4}-\dfrac{2x-5}{6}\)
(ii) \(\dfrac{5}{6a}+\dfrac{2}{9a^2}\)
(i) The LCD of \(4\) and \(6\) is \(12\). Put brackets round each numerator before you multiply.
| \(\dfrac{x+3}{4}-\dfrac{2x-5}{6}\) | \(=\) | \(\dfrac{3(x+3)}{12}-\dfrac{2(2x-5)}{12}\) |
| \(\) | \(=\) | \(\dfrac{3x+9-4x+10}{12}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{19-x}{12}}\) |
(ii) The LCD needs the LCM of the numbers, \(18\), and the highest power of \(a\), \(a^2\). So it is \(18a^2\).
| \(\dfrac{5}{6a}+\dfrac{2}{9a^2}\) | \(=\) | \(\dfrac{5\times 3a}{18a^2}+\dfrac{2\times 2}{18a^2}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{15a+4}{18a^2}}\) |
Using \(6a\times 9a^2 = 54a^3\) also works, but you then have to cancel at the end.
(i) \(\dfrac{3}{x+4}-\dfrac{2}{x-3}\)
(ii) \(\dfrac{7}{x-5}-\dfrac{4}{5-x}\)
(i) The denominators have no common factor, so the LCD is their product.
| \(\dfrac{3}{x+4}-\dfrac{2}{x-3}\) | \(=\) | \(\dfrac{3(x-3)-2(x+4)}{(x+4)(x-3)}\) |
| \(\) | \(=\) | \(\dfrac{3x-9-2x-8}{(x+4)(x-3)}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{x-17}{(x+4)(x-3)}}\) |
(ii) \(5-x\) is the negative of \(x-5\). Write \(5-x = -(x-5)\), so the second fraction changes sign.
| \(\dfrac{7}{x-5}-\dfrac{4}{5-x}\) | \(=\) | \(\dfrac{7}{x-5}+\dfrac{4}{x-5}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{11}{x-5}}\) |
Multiplying the denominators together in (ii) gives a correct but much messier answer. Spot the reversed bracket first.
Factorise the quadratic denominator. It already contains \(x-2\), so the LCD is \((x+3)(x-2)\).
| \(x^2+x-6\) | \(=\) | \((x+3)(x-2)\) |
| \(\dfrac{x+8}{(x+3)(x-2)}-\dfrac{2}{x-2}\) | \(=\) | \(\dfrac{x+8-2(x+3)}{(x+3)(x-2)}\) |
| \(\) | \(=\) | \(\dfrac{x+8-2x-6}{(x+3)(x-2)}\) |
| \(\) | \(=\) | \(\dfrac{-x+2}{(x+3)(x-2)}\) |
| \(\) | \(=\) | \(\dfrac{-(x-2)}{(x+3)(x-2)}\) |
| \(\) | \(=\) | \(\boldsymbol{-\dfrac{1}{x+3}}\) |
Always factorise the final numerator. Here \(-x+2 = -(x-2)\), and the factor \(x-2\) cancels.
(i) Show that the total time is \(\dfrac{48v}{v^2-9}\) hours.
(ii) Find the total time when \(v = 9\).
(i) Time \(=\) distance \(\div\) speed, for each leg.
| \(T\) | \(=\) | \(\dfrac{24}{v+3}+\dfrac{24}{v-3}\) |
| \(\) | \(=\) | \(\dfrac{24(v-3)+24(v+3)}{(v+3)(v-3)}\) |
| \(\) | \(=\) | \(\dfrac{24v-72+24v+72}{v^2-9}\) |
| \(\) | \(=\) | \(\dfrac{48v}{v^2-9}\) |
(ii) Substitute into the simplified form.
| \(T\) | \(=\) | \(\dfrac{48\times 9}{81-9}\) |
| \(\) | \(=\) | \(\dfrac{432}{72}\) |
| \(\) | \(=\) | \(\boldsymbol{6}\text{ hours}\) |
Check with the two legs: \(24\div 12 = 2\) h and \(24\div 6 = 4\) h, a total of \(6\) h.
Common pitfalls
Frequently asked questions
How do you add algebraic fractions with different denominators?
Find the lowest common denominator, rewrite each fraction over it by multiplying its numerator by the missing factors, then add the numerators and simplify. The denominators are never added.
How do you find the lowest common denominator of algebraic fractions?
Factorise every denominator. The LCD contains each different factor once, at the highest power it appears. For \(6a\) and \(9a^2\) it is \(18a^2\); for \(x^2-4\) and \(x+2\) it is \((x-2)(x+2)\).
Why do I get the wrong sign when subtracting algebraic fractions?
The minus sign applies to the whole second numerator. Put that numerator in brackets and multiply every term by the negative number, so \(-2(x-4)\) gives \(-2x+8\).
Should I expand the denominator in my answer?
Usually not. Leaving the denominator factorised shows whether anything cancels and makes the answer easier to check. Expand it only if the question asks for that form.
What do you do with 5 minus x in a denominator?
Write \(5-x\) as \(-(x-5)\). The fraction then changes sign and has the same denominator as any fraction over \(x-5\).
How do you simplify a fraction inside a fraction?
Simplify the top and the bottom to single fractions first. Then dividing by a fraction is the same as multiplying by its reciprocal.