Adding and Subtracting Algebraic Fractions

Try to add half a pizza to a third of a pizza. You can't just say "one half plus one third is two somethings" — halves and thirds are different-sized slices, so counting them together is meaningless. First you re-cut both pizzas into sixths — pieces of the same size — and only then can you count:

\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}

That is the whole game, and it never changes. Fractions can only be added or subtracted once they speak the same denominator — the same slice size. With numeric fractions you hunt for a common denominator you already know (like 6). With algebraic fractions there's nothing to hunt for, because you can always build one: multiply the two denominators together. The method you learned at age nine is exactly the method you'll use here — the only new skill is keeping your brackets tidy.

Get the bottoms to match, add (or subtract) the tops, keep the bottom. Everything on this page is that sentence, done carefully.

Warm-up: letters on top, numbers on the bottom

Take \frac{x}{3} + \frac{x}{4}. The letter on top changes nothing — the denominators are 3 and 4, so this is the same job as \frac{1}{3} + \frac{1}{4}. Rewrite each as an equivalent fraction over 12 (the lowest common denominator):

\frac{x}{3} + \frac{x}{4} = \frac{4x}{12} + \frac{3x}{12} = \frac{4x + 3x}{12} = \frac{7x}{12}

Multiplying top and bottom of \frac{x}{3} by 4 doesn't change its value — it's still the same fraction, just cut into smaller pieces. Once the pieces are the same size, the tops simply combine: 4x + 3x = 7x. Notice the denominator of the answer is 12, not 24 — the slice size rides along unchanged; only the numerators are added.

See it built

Watch the warm-up happen one move at a time: name the common denominator, pull each fraction onto it, then — and only then — combine the tops. Step it forwards and backwards until the rhythm feels automatic, because every harder example below follows exactly these four beats.

Notice that nothing clever happens in step three: each fraction is just multiplied, top and bottom, by whatever the other denominator is. That "multiply by the other one" move is the entire trick — and it works even when the denominators are algebra.

The rule

For any two algebraic fractions, the safe common denominator is always the product of the two denominators. Cross-multiply each numerator by the other denominator, combine the results, and put the lot over the product:

\frac{a}{p} \pm \frac{b}{q} = \frac{aq}{pq} \pm \frac{bp}{pq} = \frac{aq \pm bp}{pq}

Subtraction is identical — the minus sign just travels into the top (and, as you'll see below, it applies to the whole of the second numerator). The simplest genuinely algebraic example is two unit fractions:

\frac{1}{x} + \frac{1}{y} = \frac{y}{xy} + \frac{x}{xy} = \frac{y + x}{xy}

The denominator xy is the size of the pieces, so it rides along unchanged; only the numerators are added. Read that answer again: it is \frac{x+y}{xy}, not \frac{2}{x+y} — adding the bottoms is never allowed, and the "Watch out!" panel further down shows how to catch yourself doing it.

Full slow motion: \frac{2}{x} + \frac{3}{x+1}

Now the denominators themselves contain x. Nothing changes. Here is every step, with nothing skipped.

Step 1 — build the common denominator. Multiply the two denominators: x \times (x+1) = x(x+1). Treat x+1 as a single object — keep it in its bracket.

Step 2 — scale each fraction. Multiply top and bottom of each fraction by the other denominator:

\frac{2}{x} = \frac{2(x+1)}{x(x+1)} \qquad\text{and}\qquad \frac{3}{x+1} = \frac{3x}{x(x+1)}

Step 3 — combine the tops over the shared bottom.

\frac{2(x+1)}{x(x+1)} + \frac{3x}{x(x+1)} = \frac{2(x+1) + 3x}{x(x+1)}

Step 4 — expand and collect the numerator. 2(x+1) + 3x = 2x + 2 + 3x = 5x + 2, so

\frac{2}{x} + \frac{3}{x+1} = \frac{5x + 2}{x(x+1)}

And stop there. Expand the top, but leave the bottom factorised as x(x+1) — a factorised denominator shows at a glance whether anything cancels, and it's the form mark schemes expect. Sanity-check with a number: at x = 1 the left side is 2 + \tfrac{3}{2} = \tfrac{7}{2}, and the right side is \tfrac{7}{2} too. A ten-second check like this catches most slips.

Subtraction: escort the minus sign

Subtraction is where marks go to die — not because the method changes, but because the minus sign must hit the whole of the second numerator. Take \frac{3}{x-1} - \frac{2}{x+2}.

Common denominator: (x-1)(x+2). Scale each fraction by the other denominator and combine:

\frac{3}{x-1} - \frac{2}{x+2} = \frac{3(x+2) - 2(x-1)}{(x-1)(x+2)}

Now the dangerous move. Expand the numerator keeping the brackets until the very last moment. The second term is -2(x-1): the -2 multiplies both the x and the -1:

3(x+2) - 2(x-1) = 3x + 6 - 2x + 2 = x + 8

Note the +2: minus two times minus one is plus two. Writing 3x + 6 - 2x - 2 (getting x + 4) is the single most common error on this whole topic. The finished answer:

\frac{3}{x-1} - \frac{2}{x+2} = \frac{x + 8}{(x-1)(x+2)}

Denominator left factorised, numerator fully collected. Quick numeric check at x = 2: left is 3 - \tfrac{1}{2} = \tfrac{5}{2}; right is \tfrac{10}{1 \times 4} = \tfrac{5}{2}. It survives.

The minus sign hits the whole second numerator. When you subtract, the entire top of the second fraction goes in brackets: -(2x - 4) = -2x + 4. Forgetting to flip the second term — writing -2x - 4 — is the classic error. Write the brackets, every time, and only remove them when you expand.
Never add the denominators. \frac{1}{x} + \frac{1}{y} \ne \frac{2}{x+y}. Don't take this on trust — test it: put x = y = 2. The left side is \frac{1}{2} + \frac{1}{2} = 1; the impostor on the right is \frac{2}{4} = \frac{1}{2}. They're not even close. The true sum is \frac{x+y}{xy} (which gives \frac{4}{4} = 1. Correct.) Whenever you suspect a "rule" you've half-remembered, feed it small numbers.
Leave denominators factorised. Write x(x+1), not x^2 + x, and (x-1)(x+2), not x^2 + x - 2. A factorised bottom makes cancelling visible and later work (solving equations with fractions) far easier. Expanding it gains nothing and hides everything.

Drive to a town 60 miles away at 30 mph, then drive straight back at 60 mph. Average speed 45 mph, right? Wrong — and this page's algebra says why. The trip out takes 2 hours, the trip back 1 hour: 120 miles in 3 hours is 40 mph. You spent longer travelling slowly, so the slow leg drags the average down.

In general, covering the same distance at speeds u then v, the times are proportional to \frac{1}{u} + \frac{1}{v} = \frac{u+v}{uv} — an algebraic fraction sum! — and the true average works out as

\text{average speed} = \frac{2}{\frac{1}{u} + \frac{1}{v}} = \frac{2uv}{u+v}

This is called the harmonic mean, and it turns up wherever rates share a fixed workload: two taps filling one bath, resistors in parallel, camera f-stops. The ancient Egyptians would have approved — their scribes wrote almost all fractions as sums of distinct unit fractions like \frac{2}{7} = \frac{1}{4} + \frac{1}{28}, so adding \frac{1}{u} + \frac{1}{v}-style expressions was, for a thousand years, simply what arithmetic was.

Simplify the result

After combining, always check whether the answer can be reduced — cancel any common factor of the numerator and denominator, just as in simplifying algebraic fractions. Two quick examples:

\frac{2x}{6} + \frac{x}{6} = \frac{3x}{6} = \frac{x}{2} \qquad\qquad \frac{3}{2x} + \frac{1}{2x} = \frac{4}{2x} = \frac{2}{x}

This is another reason to keep the denominator factorised: a common factor like the 2 in 2x is easy to spot and cancel, but it vanishes from sight the moment everything is multiplied out. Most exam answers don't simplify further — but the one-line check costs nothing and occasionally saves a mark.

See it explained

Sal Khan works through adding two rational expressions whose denominators differ — building the common denominator by multiplying, scaling each numerator, then collecting. Watch for exactly the four beats from the slow-motion example above.