1
The main bar is a division
(A)/(B) means A ÷ B no matter how complicated A and B are. Once each is a single fraction, “Multiplying and Dividing Rational Expressions” finishes the job by flipping B.
Lesson · Rational and Radical Expressions · Rational expressions
Complex Fractions
Nothing new is needed here — only “Multiplying and Dividing Rational Expressions” and “Adding and Subtracting Rational Expressions” applied in one of two orders. The second order, multiplying through by the LCD of the small denominators, is usually much shorter, and both must agree.
method 1: combine the top, combine the bottom, then divide
method 2: multiply top and bottom by the LCD of the small parts
(1/x − 1/3)/(x − 3) = −1/(3x)
every small denominator contributes its own exclusion
One expression, two routes, one answer
Core ideas
What this lesson establishes
A complex fraction is not a new object. It is a division, written vertically, whose two halves happen to need simplifying first.
2
Multiplying top and bottom by the LCD is legal
It is the fundamental principle of “Simplifying Rational Expressions” again: multiplying both parts by the same non-zero expression leaves the value unchanged, and choosing the LCD clears every small denominator at once.
3
Exclusions come from every level
The small denominators, the main denominator, and any value making the whole lower half zero are all excluded. The simplified answer typically shows none of them.
The material
Two methods, and why they agree
The slow reliable route, then the fast one, on the same expression.
Definition · Complex fraction
A complex fraction is a rational expression in which the numerator, the denominator, or both, themselves contain fractions. (1/x − 1/3)/(x − 3) and (1/x + 1/y)/(1/x − 1/y) are typical.
Method 1 treats the bar as what it is. Combine the numerator into a single fraction using “Adding and Subtracting Rational Expressions”, do the same to the denominator, and then divide the one by the other using “Multiplying and Dividing Rational Expressions”. It always works and it is often long.
Method 2 removes the small fractions before they can cause trouble. Find the LCD of every small denominator that appears anywhere in the expression, and multiply both the whole numerator and the whole denominator by it.
Theorem · Why method 2 is legal
For any non-zero expression M, (A)/(B) = (AM)/(BM). Choosing M to be the LCD of the small denominators makes AM and BM free of fractions, because each small denominator divides M.
Example · The same expression, both ways
Take (1/x − 1/3)/(x − 3). Method 1: the top is (3 − x)/(3x), so the whole thing is (3 − x)/(3x) ÷ (x − 3) = (3 − x)/(3x(x − 3)). Method 2: multiply top and bottom by 3x to get (3 − x)/(3x(x − 3)) directly. Same expression, one step fewer.
Either way the finish is the sign observation from “Simplifying Rational Expressions”: 3 − x = −(x − 3), so the quotient is −(x − 3)/(3x(x − 3)), which reduces to −1/(3x).
That answer is startlingly simple, and it is where the domain discipline pays. −1/(3x) looks like it excludes only x = 0. The original excluded x = 0, because of 1/x, and x = 3, because the main denominator x − 3 was zero there. Both conditions belong to the answer.
Example · Two variables
(1/x + 1/y)/(1/x − 1/y). The LCD of the small denominators is xy. Multiplying top and bottom by it gives (y + x)/(y − x), valid when x ≠ 0, y ≠ 0 and x ≠ y — the last because the original lower half is zero when 1/x = 1/y.
That third condition is the one worth pausing on. It does not come from a denominator at all; it comes from the requirement that the thing you are dividing by is not zero. In a complex fraction the divisor is an expression, so you have to ask when that expression vanishes.
Interactive
Complex fractions
Two methods, computed separately, meeting at the same answer
Method
Clearing a complex fraction
Method 2, which is the one to reach for by default.
1
Find every small denominator
Look in the numerator and the denominator both. In (1/x − 1/3)/(x − 3) they are x and 3; the x − 3 is the main denominator, not a small one.
2
Take their LCD and record the exclusions
Here the LCD is 3x. Every small denominator's zeros are excluded, and so are the zeros of the whole lower half.
3
Multiply the whole top and the whole bottom by it
Both halves, in their entirety — brackets around each before you distribute. Every small fraction disappears in this one step.
4
Simplify what is left
You now have an ordinary rational expression. Factor, cancel, and attach the exclusions from step two.
Clearing the small fractions
Score 0 / 0Choose the answer you can justify.
Simplify (1/x − 1/3)/(x − 3)
Method 2, with the sign flip at the end and a numerical check.
(1/x − 1/3) / (x − 3)
small denominators: x and 3 LCD = 3x
exclusions: x ≠ 0 (from 1/x) x ≠ 3 (main denominator)
multiply top and bottom by 3x:
= (3x·(1/x) − 3x·(1/3)) / (3x(x − 3))
= (3 − x) / (3x(x − 3))
3 − x = −(x − 3)
= −(x − 3) / (3x(x − 3))
= −1/(3x), x ≠ 0, x ≠ 3
check at x = 1 : (1 − 1/3)/(1 − 3) = (2/3)/(−2) = −1/3
−1/(3·1) = −1/3 ✓
The answer is a single small fraction with no memory of where it came from. Nothing in −1/(3x) suggests that x = 3 is forbidden, and it is: at x = 3 the original asks you to divide by 0.
For a faded two-method check, simplify (1/x + 1/2)/(1/x − 1/2). The supplied LCD of the small denominators is 2x. First multiply the whole top and bottom by it; then restart by combining each half and dividing. Both routes should give (x + 2)/(2 − x), with x ≠ 0, 2. If 2 is missing, you checked the small denominators but never asked when the entire lower half equals zero.
Common mistakes
1
Multiplying only part of the numerator by the LCDThe LCD multiplies the whole of the top and the whole of the bottom. Bracket each half first; 3x·(1/x − 1/3) is 3 − x, not 3 − 1/3.
2
Losing the main denominator's exclusionIn (1/x − 1/3)/(x − 3) the value x = 3 is excluded even though x − 3 never appears in the final −1/(3x). It was the divisor.
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Forgetting that the lower half can be zero without a denominator being zero1/x − 1/y is zero when x = y, and dividing by it is undefined there. The rule is about the divisor's value, not about its denominators.
Check your understanding
✓
Finish when you can do one both ways and get the same thing twice.
Take any complex fraction, clear it by combining and dividing, then clear it again by multiplying through by the LCD, and confirm the two answers agree. Being able to check yourself without a solution key is the point; the two methods are independent.
Note: The lab runs both methods on whatever you type and prints the results side by side. They were computed separately, so agreement is evidence rather than decoration — and where the answers look different, they are the same expression written two ways, which is itself worth seeing.