Simplifying Algebraic Fractions

Rearrange a real formula — combining two electrical resistances, say, or two lens distances — and you often end up with a bulky fraction, a tangle of terms stacked on top of more terms. Simplifying trims it back to the cleanest equivalent expression, so it is easier to read and quicker to compute.

You already know how to do this. Hand anyone the fraction \frac{12}{18} and they'll reduce it without blinking: both 12 and 18 contain a factor of 6, so

\frac{12}{18} = \frac{6 \times 2}{6 \times 3} = \frac{2}{3}.

You didn't subtract anything, you didn't wave a wand — you spotted a number that multiplies both the top and the bottom, and divided it out. Now look at this monster:

\frac{x^2 + 5x + 6}{x^2 - 4}

It looks nothing like \frac{12}{18} — but it is exactly the same game. Factorise the top and the bottom, and a shared factor pops into view:

\frac{x^2 + 5x + 6}{x^2 - 4} = \frac{(x + 2)(x + 3)}{(x + 2)(x - 2)} = \frac{x + 3}{x - 2}.

The (x + 2) plays the part the 6 played: a whole factor multiplying the entire top and the entire bottom, so it divides out. That's the whole topic in one line: factorise first, then cancel factors — and only factors. Everything on this page is that one rule, plus the traps people fall into when they forget the second half of it.

The idea, slowly

An algebraic fraction is just a fraction whose top and bottom are expressions in x rather than plain numbers. You simplify it the very same way you reduce a numeric fraction: find a factor shared by the top and the bottom, and cancel it.

The trick with algebra is that the shared factor is usually hidden until you factorise the top and the bottom first. Take this fraction:

\frac{x^2 - 9}{x + 3}

As it stands, nothing on top matches anything below — x^2 - 9 and x + 3 look like strangers. But the numerator is a difference of two squares, so it factorises as (x + 3)(x - 3). Suddenly an (x + 3) sits in both the top and the bottom — a common factor we can cancel:

\frac{x^2 - 9}{x + 3} = \frac{(x + 3)(x - 3)}{x + 3} = x - 3

Why is that legal? Because cancelling is just dividing the top and the bottom by the same thing. Peel the fraction apart and you can watch the factor turn into a harmless 1:

\frac{(x + 3)(x - 3)}{x + 3} = \frac{x + 3}{x + 3} \times (x - 3) = 1 \times (x - 3) = x - 3.

See it built

Here is that same simplification as a four-step film. Notice that nothing gets crossed out until the top has been factorised — the shared (x + 3) simply isn't visible before then. That's the rhythm to internalise: factorise, spot, cancel, read off. Step through it.

Worked examples

Example 1 — monomials: numbers and letters cancel separately. Simplify \frac{6x^3}{9x}. Deal with the numbers first, then the powers of x:

\frac{6x^3}{9x} = \frac{6}{9} \times \frac{x^3}{x} = \frac{2}{3} \times \frac{x \cdot x \cdot x}{x} = \frac{2x^2}{3}.

The 6 and 9 share a factor of 3; the x^3 and x share a factor of x. Two little cancellations, done in your head one at a time.

Example 2 — factorise both, top and bottom. Simplify \frac{x^2 + 5x + 6}{x^2 + 7x + 12}. Neither is factorised yet, so we do both. Top: two numbers that multiply to 6 and add to 5 are 2 and 3. Bottom: multiply to 12, add to 7 — that's 3 and 4. So:

\frac{x^2 + 5x + 6}{x^2 + 7x + 12} = \frac{(x + 2)(x + 3)}{(x + 3)(x + 4)} = \frac{x + 2}{x + 4}.

The (x + 3) cancels; the (x + 2) and (x + 4) stay put, because nothing matches them. Do not be tempted to "cancel the x's" in \frac{x + 2}{x + 4} — that fraction is finished.

Example 3 — when nothing cancels, stop. Simplify \frac{x + 2}{x + 3}. Both top and bottom are already fully factorised — each is a single factor — and they are different factors. There is no shared factor, so there is nothing to do: \frac{x + 2}{x + 3} is already in its simplest form. Knowing when you're done is a skill in itself; an answer that "looks unfinished" is often perfectly finished. (If you're ever unsure, test with a number: at x = 1 the fraction is \frac{3}{4}, and no amount of "cancelling the x's" — which would claim \frac{2}{3} — can change that.)

Every year, in every exam hall, someone writes \frac{x + 6}{6} = x. It feels so tidy — a 6 up top, a 6 below, slash them both. But it's completely wrong, and one test exposes it instantly: try x = 1. The left side is \frac{1 + 6}{6} = \frac{7}{6}; the right side is 1. Not equal — the "cancellation" was a fake.

The same disease in another outfit: \frac{x^2 + 5}{x^2} \neq 5. At x = 1 the left side is 6, not 5. Busted again.

Why do these fail? Because the 6 in x + 6 is a term (something added), not a factor (something multiplying the whole expression). Cancelling means dividing top and bottom by the same thing — and dividing x + 6 by 6 gives \frac{x}{6} + 1, not x. Compare the legal move: in \frac{6x}{6} the 6 multiplies everything on top, so \frac{6x}{6} = x is fine.

The rule of thumb: you may only cancel something that multiplies the entire top and the entire bottom. If there's a + or - loose on either level, factorise fully first — and when in doubt, plug in x = 1 and see whether both sides agree.

Here is a fraction simplified by pure crime: \frac{16}{64}. Cancel the sixes! — \frac{1\cancel{6}}{\cancel{6}4} = \frac{1}{4}. And the outrageous thing is… \frac{16}{64} really does equal \frac{1}{4}. The method is nonsense, the answer is right.

These flukes are called anomalous cancellations, and among fractions of two-digit numbers (with different digits) there are exactly four of them:

\frac{16}{64} = \frac{1}{4}, \quad \frac{19}{95} = \frac{1}{5}, \quad \frac{26}{65} = \frac{2}{5}, \quad \frac{49}{98} = \frac{4}{8}.

So why is the method still wrong? Because the 6 in 16 isn't a factor of it — 16 means 10 + 6, a sum, and digits are just addresses in that sum. The real reason \frac{16}{64} = \frac{1}{4} is the honest cancellation \frac{16 \times 1}{16 \times 4}. Try the digit trick on almost anything else — say \frac{12}{24}, "cancel the 2s" to get \frac{1}{4} — and it collapses (the true value is \frac{1}{2}). A method that works four times out of thousands isn't a method; it's a lottery ticket. Algebra runs on reasons, not flukes.

A quiet footnote: the forbidden value

We happily wrote \frac{x^2 - 9}{x + 3} = x - 3. But look what happens at x = -3: the original fraction becomes \frac{0}{0} — division by zero, which is meaningless — while x - 3 cheerfully gives -6. The two expressions agree at every other value of x, but at the root of the factor we cancelled, the original was never defined at all.

Graphically, y = \frac{x^2 - 9}{x + 3} is the straight line y = x - 3 with a single point punched out — a hole at (-3, -6):

At GCSE you won't usually be asked to write it, but it's worth knowing: cancelling a factor doesn't un-forbid its root. Strictly, \frac{x^2 - 9}{x + 3} = x - 3 for x \neq -3. File it away — it becomes important later, when you meet limits in calculus, where "what happens right next to the hole" is the whole story.

See it explained

For a second voice on the same idea, here is Sal Khan factorising the top and bottom of rational expressions and cancelling the shared factors to write them in lowest terms — including a careful note on exactly the forbidden-value point above. Watch how every single simplification starts the same way: factorise first.