Multiplying both sides by a common denominator turns an equation full of fractions into one with none. The move is the multiplication property with a convenient constant, so it changes nothing about the answer — provided the multiplier reaches every term on both sides.
Core ideas
Nothing here is a new rule about fractions. It is one application of the multiplication property, chosen so that the arithmetic afterwards is easy.
Multiplying both sides by a nonzero number produces an equivalent equation. The LCD is just a well-chosen value of that number, and the equation you get has exactly the same solutions as the one you started with.
Including the terms that have no denominator. Multiplying x/2 + x/3 = 5 by 6 and leaving the 5 alone gives 3x + 2x = 5, an equation about something else entirely.
Any common multiple of the denominators clears them. 12 works on x/2 and x/3 as well as 6 does; it merely doubles every number in the rest of the problem. Choosing the least one is a habit, not a law.
The material
Finding the multiplier, applying it to every term, and the two places the brackets matter.
The least common denominator of a set of fractions is the smallest positive number that every denominator divides exactly — the least common multiple of the denominators.
For 2 and 3 it is 6. For 4 and 6 it is 12, not 24: 24 is the product, and the product is a common multiple but not usually the least one.
For small denominators, list multiples until the lists first meet: the multiples of 4 are 4, 8, 12, ... and of 6 are 6, 12, ..., so the LCD is 12. For a longer list, prime-factor each denominator and take every prime at the highest power that occurs:
The list method is quickest for two small denominators; the prime-power method scales. In either case, verify the result by dividing it by every denominator before multiplying the equation.
Using 12 instead gives 6x + 4x = 60, so 10x = 60 and x = 6. The same answer, reached through larger numbers. That is the only penalty for not using the least common denominator.
(x + 1)/3 is the whole of x + 1 divided by 3. When the multiplier arrives, the bracket has to stay:
Two things happen in the expansion and both are worth naming. 12 · (x + 1)/3 is 4(x + 1), which is 4x + 4 — not 4x + 1. And −3(x − 2) is −3x + 6, because the minus reaches both terms inside the bracket.
0.2 is 1/5, 1.5 is 3/2 and 3.1 is 31/10. Multiplying by a power of ten clears them all at once: use 10 for one decimal place, 100 for two, and so on for the longest one present.
The multiplier here is a fixed nonzero number, so the move is reversible: divide by it and you are back where you started. The solution set therefore cannot change, and no candidate needs to be thrown out afterwards.
That guarantee depends on the multiplier being a number. When a denominator contains the variable — 3/(x − 2) — the multiplier is zero for one value of x, the move is not reversible there, and a candidate can appear that solves nothing. “Ratio, Proportion and Percent” meets the first case of that, and Rational and Radical Expressions treats it properly.
x/6 + x/3 = x/2, multiplied through by 6, becomes x + 2x = 3x, that is 3x = 3x. Every x cancels and 0 = 0 is left. The two sides were the same expression written differently — x/6 + x/3 = x/6 + 2x/6 = 3x/6 = x/2 — so every number is a solution. “Identities and Contradictions” names this outcome.
Interactive
One multiplication, every term
Method
Find the multiplier first, then use it on everything without exception.
0.2 counts as 1/5. Write the denominators down rather than holding them in your head; the list is what the next step works from.
That is the LCD. Any common multiple works, so if the least one is not obvious, the product of the denominators is a safe fallback that costs only larger arithmetic.
Count the terms first and count them again afterwards. 12 · (x + 1)/3 is 4(x + 1); the bracket is not optional, and neither is multiplying the terms that had no denominator.
The check goes back into the equation with the fractions in it. That is the one that was asked, and it is the only version that can confirm the clearing step itself.
The first has multi-term numerators and a subtraction; the second is decimals.
Three terms went into the multiplication in A and three came out. That count is the fastest guard against the standard error, which is multiplying the fractions and leaving the 1 on the right untouched.
The sign on −3(x − 2) is the second trap. It produces +6, and writing −6 instead gives x = 14 — an answer that survives every later step and fails the check.
For a faded clearing, use (2x − 1)/6 + (x + 2)/4 = 3. The supplied LCD is 12 and the cleared line is 2(2x − 1) + 3(x + 2) = 36. Expand both brackets, solve for an exact fraction, and check it in the original fractional equation before asking the lab.
Without the lab, solve 0.3x − 1.2 = 0.05x + 0.8. State a positive power-of-ten multiplier, apply it to all four written terms, solve, and check in the decimal original. Then find the LCD of 6, 8 and 12 by prime powers and verify that each denominator divides it.
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