Here is a recognition pattern for any algebraic expression that contains fractions with different denominators. The moment your eye lands on something like \tfrac{x}{2} + \tfrac{x}{3} or \tfrac{x+1}{3} - \tfrac{2x - 1}{5}, stop. Do not try to combine the numerators yet. The first move is always the same: find the common denominator, rewrite each fraction to that denominator, then combine. Students who skip this step almost always end up adding denominators — \tfrac{1}{2} + \tfrac{1}{3} = \tfrac{2}{5} — which is the single most famous fraction mistake in school arithmetic, and it migrates wholesale into algebra if you let it.

This is a structural rule, not a personal preference. Different denominators are a signal that the pieces are not yet countable together. Common denominator first, combination second. Get that order wrong and you will lose every sign and magnitude in the problem.

Why you can't combine across different denominators

Go back to pure arithmetic for a moment. You already know that \tfrac{1}{2} + \tfrac{1}{3} is not \tfrac{2}{5}. Why not? Because the denominator is telling you the size of the pieces. A half and a third are different-sized pieces, and you cannot add "one piece" to "another piece" unless the pieces are the same size. Convert first: \tfrac{1}{2} = \tfrac{3}{6} and \tfrac{1}{3} = \tfrac{2}{6}. Now both fractions are measuring in sixths, and \tfrac{3}{6} + \tfrac{2}{6} = \tfrac{5}{6} — five sixths of one whole, which is the actual answer.

The same logic applies with variables. \tfrac{x}{2} means "half of some quantity x" and \tfrac{x}{3} means "a third of that same x." These are not the same-sized slice of x, and you cannot just call their sum \tfrac{2x}{5} and walk off. The denominators count pieces, not numerators. Add them blindly and you change the value.

Quick sanity check: at x = 6, \tfrac{x}{2} + \tfrac{x}{3} = 3 + 2 = 5. If you believed \tfrac{2x}{5}, you would get \tfrac{12}{5} = 2.4. Two different numbers, ten-second check, done.

The common-denominator algorithm, one more time

Since the rule gets forgotten, here it is plainly. Given fractions \tfrac{a}{p} + \tfrac{b}{q} - \tfrac{c}{r} (or any combination):

  1. Find the LCM of all denominators. Call it L.
  2. For each fraction, multiply its numerator by \tfrac{L}{\text{its denominator}}.
  3. Put every rewritten numerator over the single denominator L.
  4. Simplify the combined numerator like any ordinary algebraic sum.

Every fraction goes through steps 2 and 3 — no exceptions. Start with three fractions, end with one fraction whose numerator has three contributions.

Two different-sized grids rewritten to a common denominator gridOn the left, a grid of 2 cells with one cell shaded, labelled one half. Next to it, a grid of 3 cells with one cell shaded, labelled one third. An arrow points to the right to a grid of 6 cells with 3 cells shaded and then 2 more cells shaded, labelled three sixths plus two sixths equals five sixths. different-sized pieces cannot be added until they are the same size 1/2 halves + 1/3 thirds rewrite 3/6 + 2/6 = 5/6 sixths — same size, now addable the common denominator rewrites every fraction in the same units only then do the numerators stack into a single sum
Halves and thirds are not the same-sized piece — you cannot stack them. Rewriting each to sixths, the common denominator, puts everything in matching units, and only then does $3/6 + 2/6 = 5/6$ make sense. The exact same reasoning carries over to $x/2 + x/3$ with $x$ in the numerators.

Worked example 1: the clean case

Simplify \tfrac{x}{2} + \tfrac{x}{3}.

Step 1: LCM of 2 and 3 is 6.

Step 2: rewrite each fraction to denominator 6. \tfrac{x}{2} = \tfrac{3x}{6} (multiply top and bottom by 3). \tfrac{x}{3} = \tfrac{2x}{6} (multiply top and bottom by 2).

Step 3: put both over the single denominator. \tfrac{3x}{6} + \tfrac{2x}{6} = \tfrac{3x + 2x}{6} = \tfrac{5x}{6}.

Why: the conversions \tfrac{x}{2} = \tfrac{3x}{6} and \tfrac{x}{3} = \tfrac{2x}{6} each multiply top and bottom by the same number, which does not change the value of the fraction. Once both fractions are measured in sixths, the numerators can be added directly.

Check at x = 6: original is 3 + 2 = 5, simplified is \tfrac{30}{6} = 5. Matches.

Worked example 2: signs, signs, signs

Simplify \tfrac{x + 1}{3} - \tfrac{2x - 1}{5}.

Step 1: LCM of 3 and 5 is 15.

Step 2: rewrite. \tfrac{x + 1}{3} = \tfrac{5(x + 1)}{15} and \tfrac{2x - 1}{5} = \tfrac{3(2x - 1)}{15}.

Step 3: combine into a single fraction over 15.

\frac{x+1}{3} - \frac{2x-1}{5} \;=\; \frac{5(x+1) - 3(2x-1)}{15}.

Step 4: expand the numerator carefully. This is the step where most sign errors appear. The minus sign in front of 3(2x - 1) distributes all the way through, so -3(2x - 1) = -6x + 3, not -6x - 3.

5(x + 1) - 3(2x - 1) \;=\; 5x + 5 - 6x + 3 \;=\; -x + 8.

Why: inside the second bracket, -3 \cdot 2x = -6x and -3 \cdot (-1) = +3. The second product is positive because the negative from the coefficient meets the negative on the -1. If you wrote -6x - 3, you kept the -1 as -1 and forgot it picks up the sign of the coefficient. This is the exact distribute-the-minus habit from the coefficient-distribution page.

Final answer: \dfrac{-x + 8}{15}, which you can also write as \dfrac{8 - x}{15}.

Check at x = 1: original is \tfrac{2}{3} - \tfrac{1}{5} = \tfrac{10}{15} - \tfrac{3}{15} = \tfrac{7}{15}. Simplified is \tfrac{-1 + 8}{15} = \tfrac{7}{15}. Matches.

Worked example 3: when the denominators are expressions

Simplify \dfrac{2}{x} + \dfrac{3}{x + 1}.

The denominators are not numbers — they are x and x + 1. The algorithm does not care. The LCM of x and x + 1 is simply their product x(x + 1), because x and x + 1 share no common factor.

Step 2: rewrite. \tfrac{2}{x} = \tfrac{2(x + 1)}{x(x + 1)} and \tfrac{3}{x + 1} = \tfrac{3x}{x(x + 1)}.

Step 3: combine.

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

Final answer: \dfrac{5x + 2}{x(x + 1)}.

Check at x = 2: original is \tfrac{2}{2} + \tfrac{3}{3} = 1 + 1 = 2. Simplified is \tfrac{10 + 2}{2 \cdot 3} = \tfrac{12}{6} = 2. Matches.

The sign trap

Worked example 2 was not a coincidence. Whenever you subtract a fraction whose numerator has more than one term, the minus sign distributes to every term in that numerator. Every one. Not just the first. This is the same trap as -(2x - 1) = -2x + 1, not -2x - 1. If you routinely forget to flip the second term, you will routinely get these problems wrong by a constant.

The safest habit is to put brackets in explicitly the moment you pull the numerators over the common denominator. Write \tfrac{5(x + 1) - 3(2x - 1)}{15} with brackets visible, then expand in a separate step. The brackets tell you the minus sign applies to the whole numerator, not just the leading term.

Cross-multiply only for equations

A common confusion: "can't I just cross-multiply?" No. Cross-multiplication is an equation move — the shortcut for when two fractions are equal, like \tfrac{a}{b} = \tfrac{c}{d} \Rightarrow ad = bc. That shortcut only works because there is an equals sign you can multiply both sides of.

An expression has no equals sign. \tfrac{a}{b} + \tfrac{c}{d} is a single object, not two sides of an equation. You cannot "multiply both sides" because there are no two sides. The only way to combine these fractions is the common-denominator rewrite. Cross-multiplying on an expression quietly changes its value and leaves you with something unrelated to what you started with.

Recognise sub-patterns

A bit of pattern-spotting saves time. When the denominators are conjugates — (x + 1) and (x - 1), or (x + a) and (x - a) — their LCM is their product, and that product collapses via the difference of squares.

\frac{x + 1}{x - 1} + \frac{x - 1}{x + 1}.

Denominators are (x - 1) and (x + 1). LCM is (x - 1)(x + 1) = x^2 - 1. Rewrite:

\frac{(x + 1)^2 + (x - 1)^2}{x^2 - 1} \;=\; \frac{(x^2 + 2x + 1) + (x^2 - 2x + 1)}{x^2 - 1} \;=\; \frac{2x^2 + 2}{x^2 - 1}.

Recognising the conjugate denominators up front means you do not have to multiply (x - 1)(x + 1) out by hand — you already know it is x^2 - 1. Small savings, but they add up over an exam.

Close

When you see fractions with different denominators in an expression, the order of operations is rigid: LCM first, rewrite second, combine third. Skip the LCM and you will end up adding denominators or subtracting without distributing the minus, and either way the final answer will be wrong by exactly the amount that makes the mistake hard to catch. Common denominator first, combination second. That order is not optional — it is what the arithmetic actually asks of you.